How do i solve by using the quadratic formula?\n5x2 – x + 12 = 0
206019
x=-(-1)+/_ {(-1)^2 -4*5*12} all over 2*5\n= 1+/- {1-240} over 10\n= 1+/- i{249} over 10\n\n{}= square root, ^= to the power of,\n\nif you need more help let me know via email.
552613
Use quadratic formula:\nx^2 - 2x + 1 = -2 + 1\n(x-1)^2 = -1\nso no roots so no critical points
658238
You know the first term is x^2 so you can write:\n(x     )(x     )\nYou know that the second term is -4x:\nIf factors are of the form (x+A)(x+B) then  Ax+Bx = -4x\nYou also know that the last term is -5, so A*B=-5\n\nJust find the values of A and B that satisfy these equations:\nAx+Bx = -4x\nA*B = -5\n\nA = -5 and B = 1\n\nNOW that we have the factors\n(x-5)(x+1)=0\nTo get 0 either section of the left hand side can be 0, so solve both cases:\nCASE 1:\nx-5 =0\nx = 5\nCASE 2:\nx+1=0\nx=-1\n\nANSWER x=5 or x=-1
402841
Eliminate the y to get\n2-x^2=2x+4\nx^2+2x+2=0\nusing quadratic formula\n(-2+-sqrt(4-4*2))/2\nwe see there are no real roots: the two functions do not intersect\nx= -1+-i
376501
fistly put the x's on the same side so we weill have the original eqn being 12x - 4 = 7x - 29\n                  12x - 7x - 4 = 7x - 7x - 29\n                 12x - 7x -4 +4 = 4 - 29\n                  12x - 7x = 4 - 29\n                    5x = -25\n         Divide both sides by 5 and we get x = -5
158061
Here is how I would approach the problem:\n\nTake the equation and put it in the form of a quadratic with X as the variable:\n\n5*x^2 + y^2 - 4*xy - 8*x + 2*y + 5 = 0\n\n5*x^2 + (-4*y - 8)x + (y^2 + 2*y + 5) = 0\n\nThen for the form ax^2 + bx + c\n\na= 5\n\nb= -(4*y + 8)\n\nc= y^2 + 2*y + 5\n\nThe solution x = (-b +/- sqrt(b^2 - 4*a*c))/ 2*a\n\nrequires for all REAL values (not imaginary),\n\nthat b^2 - 4*a*c >= 0\n\n    >= greater than or equal to\n\nsubstituting from above\n\n-(4*y + 8)^2 - 4*5*(y^2 + 2*y + 5) >= 0\n\n16*y^2 + 64*y + 64) - (20*y^2 - 40*y - 100) >= 0\n\n-4*y^2 + 24*y - 36 >= 0\n\ny^2 - 6*y + 9 <= 0\n\n(y - 3)^2 <= 0\n\nSince a square term will always be positive, and cannot be less than 0,\n\nThe only viable solution is\n\n(y - 3) = 0,\n\nand y = 3  \n\nThe substituting for y in the original equation\n\n5*x^2 - 20*x + 20 = 0\n\nx^2 - 4*x + 4 = 0\n\n(x-2)^2 = 0\n\nx -2 = 0\n\nx = 2\n\n\nTry using the quadratic equation to eliminate unacceptable answers.\n\nPatience and perseverence! I honestly didn't know for sure when I started that I would reach a concrete solution this way, but I had to start, and continue until i hit a roadblock or I finished. Today I got lucky, and you can too.
529621
To find the equation, just substitute the 2 roots to the "Quadramagic formula" (...)\nx² - (p + q)x + pq = 0\nwhere p and q are the roots.\n\nHence, p = 7 + i and q = 7 - i.  We substitute:\nx² - [(7 + i) + (7 - i)]x + (7 + i)(7 - i) = 0\n\nWe add and multiply\nx² - (14)x + (7² - i²) = 0\n\nSince i² = -1, then\nx² - 14x + 49 - (-1) = 0\n\nthen,\nx² - 14x + 49 + 1 = 0\n\nTherefore, the quadratic equation is\nx² - 14x + 50 = 0\n\nTherefore, the correct answer is 1).\n\n^_^
173577
Try to solve the eqs with quad rule:\na*x^2 + b*x + c=0\nx=(-b +- Sqrt(b^2-4*a*c))/(2*a)\n\nx^2+2x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+3=(x+3)*(x+1)=0; \nSo, yes, this can be factorized. \nSolutions are x=-3 and x= -1\n\nx^2+2x-3=(x+3)*(x-1)=0;\nSo, yes, this can be factorized. \nSols are x= -3 and x=1
375947
the quadratic formula is a useful tool in solving an equation involving a variable and its square without breaking it down into smaller pieces. It can only be used when it is in the\nax^2+bx+c=0  format. (a, b, and c can be negative.)\n\nIn this problem, a=2, b=-2, and c=-1. The equation for the quadratic is:\n\nx= (-b+/- sq.rt (b^2-4ac))/2a\n\nSolving for the equation given, you get:\n\nx= (2 +/- sq.rt.(4-4*2*-1))/4\n   = (2+/- sq.rt 12)/4\n   = 1.366, 0.366
445734
x^2 + 4x + 4 = 7 (subtrat 7 from both sides)\n\nx^2 + 4x -3 = 0 (use the quadratic equation formula)\n\n-b + or - the square root of (b^2 - 4ac) / a^2\n\n-4 + or - the square root of 16 - (4) (1) (-3) / 2\n\n-4 + or - the square root of 28 / 2\n\n-4 + or - (the square root of 4) (the square root of 7) / 2\n\n-4 + or - 2 (the square root of 7) / 2 (simplify by dividing 2 into  the -4 and 2 on the top to get rid of the 2 on the bottom)\n\nthat gives you x = - 2 + or - the square root of 7.
39161
U need to use the quadratic formula.
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
534746
y=5x+2\ny=5(2)+2\ny=10+2\ny=12
348539
20x^2+60x + 45\n= 5(4x^2+12x + 9) \n= 5((2x)^2+2(2x)3+3^2) = 5(2x+3)^2\n\n3x^2+12x+12 = 0\n=> x^2+4x+4 = 0 (divide by 4)\n=>(x+2) ^2 = 0 or x = -2, -2 (double root)\n\n7x^2-28 =0\n= x^2 = 4\n=> x = 2 or -2\n\narea = xy = 5*2 = 10cm^2
427163
(12x^2 +16x) - (12x-7)- (4x-5)\n12x^2 +16x -12x +7 -4x +5\n\n12x^2+12\n\nthat is what I got....\nI think you might be able to put 12 on the other side of the = sign and turn it into 12x^2 = -12 and then since both sides have 12 in common you can divide both sides by 12 and you will have x^2 = -1 then you can add 1 to both sides and you will have x^2 +1 but I think I'm getting off track by that point....it's been a while since I had to do this stuff.
295340
x = (-b +- sqrt(b^2-4ac))/(2a)\n\nThis equation solves the quadratic equation, where\n\nax^2 + bx + c = 0.
496306
5x^2 and - 50x both can be divided by 5 soooo..do this\n\n5(x^2 - 10x) \n\nthen put a space to make in a quadratic equation.  Also, put the +129 on the outside of the paranthesis\n\n5(x^2 - 10x + __ ) + 129 - __\n\ntake half of 10 which is 5.  5^2 and put it in the blank spot\nREMEMBER PUT THE OPPOSITE OF WHAT YOU PUT IN THE QUADRATIC EQUATION ON THE OUTSIDE AND YOU NEED TO MULTIPLY THE OUTSIDE BY THE 5 IN FRONT OF THE EQUATION (25 * 5)= 125\n\nso you get\n5(x^2 - 10x + 25) +129 -  125\nthen factor\n\ny= 5(x - 5)^2 + 4