How would i solve this by using the quadratic formula?    \nx2 – 15x + 54 = 0
208950
When you have an equation in the form\nax^2 + bx + C = 0\n\nyou just substitute into the quadratic formula\n(-b +/- Sqr.Rt. of (b^2 - 4ac))/2a \n\nSo for your problem:\n(15 +/- Sqr.Rt. of (15^2 - 4(1)(54)))/2(1)\n\nYou can do the arithmetic
378565
x^2 + 4x + 4=7  (first put it in the form ax^2 + bx + c = 0, by subtracting 7 from both sides of the equation)\n\nx^2 + 4x - 3 = 0   (this is called a quadratic equation)\n\nNow substitute a =1, b = 4, c = -3 in the following formula\n\nx equals -b plus or minus the square root of (b^2 - 4ac) all over 2a\n\nBecause you are solving for x^2 you expect to get 2 answers.  The plus or minus in the middle of the formula gets you 2 answers as you solve the formula for both plus and minus the square root of (b^2 minus 4ac).\n\n[If you're going to have to do this in an exam, it's probably worth learning the formula off by heart.  It's boring to have to do so but you'll probably have more confidence in the exam if you see an quadratic equation and already know the formula.]\n\nGood luck for the working out - you can do this bit!
106893
Solve by substitution.\nSolve for y^2 in the second equation by subtracting x^2 and 2x from both sides.\nx^2 + y^2 + 2x - x^2 - 2x = 19 - x^2 - 2x\ny^2 = 19 - x^2 - 2x  Now substitute this into the first equation in place of y^2\nx^2 + 2(19 - x^2 - 2x) = 33  Distribute the 2\nx^2 + 38 - 2x^2 - 4x = 33  Combine x^2 and -2x^2 and subtract 33 from both sides\n-x^2 - 4x + 5 = 0  Multiply both sides by negative one (I just do this because it is easier to factor when the x^2 is positive)\nx^2 + 4x - 5 = 0  Now factor\nWe are looking for factors of -5 that add up 4.  5 and -1 work because 5*(-1) = -5 and 5 - 1 = 4\nTherefore x^2 + 4x -5 = 0 becomes\n(x+5)(x-1) = 0  Set each part equal to 0\nx+5 = 0 or x-1 = 0\nTherefore,\nx = -5 or x = 1\nNow substitute each of these values into the y^2 equation from above.\ny^2 = 19 - x^2 - 2x\nFor x = -5\ny^2 = 19 - (-5)^2 - 2(-5)\ny^2 = 19 - 25 + 10\ny^2 = 4  Take the square root of each side to get\ny = -2 or 2\n\nFor x = 1\ny^2 = 19 - (1)^2 - 2(1)\ny^2 = 19 - 1 - 2\nY^2 = 16  Take the square root of each side to get\ny = -4 or 4\n\nThus, the solutions are\n(-5,-2), (-5,2), (1,-4), (1,4)\n\nIt is always good to check your answers, especially on a test.  To check your answers just plug each one into each equation to see if you get a true statement.
551227
well, take out 2x...2x(x-4)=0\nsolve 2x=0 and x-4=0\nx=0 or x=4
651316
Set 16t^2 +32t -225=0\n\nUse quadratic formula gives t = 2.9sec
550889
use the quadradic formula then:\n\ny^2 + y - 30 = 0\n\n(-1 +/- sqrt(1- 4 (-30)) )/ 2\n\ny=(-1 +/- 11 ) 2\n\n-6, 5 are your two answers. 10 points please ! :)
338881
the equation is:\n       55x = 224 - 59\n         55x =   165\n             x = 165/55\n               = 3
382504
You need to use the quadratic equation as it is not factorable into rational expressions.\nThe form of the equation is:\nax^2 + bx + c = 0 \nSo in your case,\na = -2, b = 6, c = -3\nand the solution is:\nx = [ -b ± √(b^2 - 4ac) ] / (2a)\n\nso substitute and reduce.\n\nx = [ -6 ± √(6^2 - 4(-2)(-3)) ] / (2(-2))\n\nx = [ -6 ± √(36 - 24) ] / (-4)\n\n= [ -6 ± √12 ] / (-4)\n= [ -6 ± 2√3 ] / (-4) \n= (3 ± √3)/2
365835
y = ax^2 + bx + c\n2 = a(0)^2 + (0)x + c\nc = 2\n\ny = ax^2 + bx + 2\n\n54 = a(70)^2 + b(70) + 2\n54 = a(4900) + b(70) + 2\n4900a + 70b = 52\n\n445 = a(42)^2 + b(42) + 2\n445 = a(1764) + b(42) + 2\n1764a + 42b = 443 \na = (443 - 42b) / 1764\n\n4900((443 - 42b) / 1764) + 70b = 52\n1230.555 - 116.666b + 70b = 52\n1230.555 - 46.666b = 52\n-46.666b = -1178.555\nb = 25.255\n\na = (443 - 42(25.255)) / 1764\na = -0.35\n\ny = (-0.35)x^2 + (25.255)x + 2
369458
I got a different answer than you.  Let's do this together:\n\nRecall that for an equation ax² + bx + c = 0, that the quadratic formula is:\n\nx = [ -b ± √ (b² - 4ac) ] ÷ 2a\n\nSo we want to find the zeros of x² + 10x + 23\n\nSo, x = [ -b ± √ (b² - 4ac) ] ÷ 2a\n\n= [ -10 ± √ (10² - 4*1*23) ] ÷ 2*1\n\n= [ -10 ± √ (100 - 92) ] ÷ 2\n\n= [ -10 ± √8 ] ÷ 2\n\n= [ -10 ± 2√2 ] ÷ 2\n\n= -5 ±√2\n\nHope that helped.\n\n~ ♥ ~
607319
I'm reading the equation as \n\n4x^2 + 8x -1 = 0\n\nUsing quadratic formula x = (-8 +/- sqrt(64+16))/8 \n\nso x = -18/8 or 2/8 = -9/4 or 1/4\n\nThose are the roots.\n\nSubstitute each of those values as y into the second formula.
436888
letus take all terms on LHS to RHS SO WE WILL GET\n23x^2-16x+2x+3-18=0\n 23x^2-14x-15=0\n \n\nhere a=23 b=(-14)  c=(-15)\n find D= b^2-4ac\n\nthen x={ (-b)+(square root of D)}/2a\n\nor x={ (-b)-(square root of D)}/2a
496696
2x(x - 2) = 3(x + 10)\n2x² - 4x = 3x + 30\n2x² - 7x - 30 = 0\n(2x + 5)(x - 6) = 0\nx = -5/2 or 6 ← option d
661990
>< you really should read the answers for your other questions and learn... hope you know this afterwards!\n\n\n6x-y=-4\n2x+2y=15\n\nuse second equation to find out what y is\n\nx+y=7.5 (divide everything by 2)\ny=7.5-x\n\nsubstitue 7.5-x as y in the first equation, find x\n\n6x-(7.5-x)=-4\n6x-7.5+x=-4\n7x=3.5\nx=0.5\n\nput in value of x in an equation and find y\n\n6(0.5)-y=-4\n3-y=-4\n-y=-7\ny=7\n\ncheck answers (use equation that was not used to find y):\n2(0.5)+2(7)=15 :)\n\nx=0.5, y=7\n\n\nx+y=1\n2x-y=-2\n\nuse first equation to find y (in relation to x) --- can find x too, but you have to plug in answer into second equation, and y is easier to deal with in 2nd equation...\n\ny=1-x\n\nput 1-x as value of y in the second equation\n2x-(1-x)=-2\n2x-1+x=-2\n3x=-1\nx=-1/3\n\nput in value of x into an equation to find y\n(-1/3)+y=1\ny=4/3\n\ncheck answers:\n2(-1/3)-(4/3)=-2 :)\n\nThe answers are x=-1/3, y=4/3\n\n\n\nthird problem:\n5x-3y=-11\nx-2y=2\n\nuse second equation to find out value of x in relation to y\nx=2+2y (add 2y to both sides to isolate x)\n\nput in that value as x for the first equation:\n5(2+2y)-3y=-11\n10+10y-3y=-11\n7y=-21\ny=-3\n\nsubstitute value of y in x-2y=2 to find x\nx-2(-3)2\nx+6=2\nx=-4\n\ncheck answers\n5(-4)-3(-3)=-11 :) answers match up\n\nhope that helped you learn how to solve this type of algebra questions!
416332
x³ - 16x\n x(x² - 4²)\n x(x - 4)(x + 4) =0\n\n1. x = 0\n2. (x-4) =0   ::  then  ::   x=4\n3. (x+4) =0   ::  then  ::   x=(-4)
414867
That's the quadratic formula. It is the solution to equations of the form\nax^2+bx+c=0
627809
I used the quadratic formula\n\na = 5\nb = -7\nc= = -1\n\n-(-7) +- sqrt[ (-7^2 - 4(5)(-1)]\n----------------------------------------\n             2(5)\n\n7 +- sqrt [ 49 - (-20)]\n---------------------------\n             10\n\n7 +- sqrt [69]\n------------------\n         10\n\n\nFor your second problem, one of the solutions is -1/3.  The other is also 4\n\nfrom (11+13) / 6 = 4 or yours (11-13) / 6 = -1/3