Solve using the quadratic equation: x2 – 3x – 23 = 5?
103889
x^2-3x-23=5\n or x^2-3x-23-5=0\nx^2-3x-28=0\n\nFor any quadratic equation:-\n\nx= +/- (sqrt. b^2- 4*a*c) / 2*a\n\n\nx= +/- ( sqrt. 9+27*4)/ 2\n x = +/- ( 11)/2\n\nAns= +11/2; -11/2
516994
The trick is to substitute a known true equation. I used 2+3=5, but any true equation will work.\n\nSubstitututing in the first equation you get\n4(2) - 3(2) + 4(3) - 3(3) = 4(5) - 3(5)\n\nwhich is 8 - 6 + 12 - 9 = 20 - 15\n\nor 5 = 5.\n\n\nIn the second equation you get\n4(2) + 4(3) - 4(5) = 3(2) + 3(3) - 3(5)\n\nwhich is 8 + 12 - 20 = 6 + 9 -15\n\nor 0 = 0.\n\n\nHence, a + b - c = 0.  In the next equation, you are dividing by zero, which is not permissible in math.  If a +  b - c had not been equal to zero, then this would have been possible.
162979
From y=-3x-3 and y=x+5 you have -3x-3=x+5 so\n-3x-x=5+3\n-4x=8\nx=-2\nNow y=x+5=-2+5=3\nThe solution is (-2,3).
1108240
2(x-5)^2 = 3\ndivide by 2:  (x - 5)^2 = 3/2\nsquare root: (x - 5) = +/- root(6)/2\nadd 5: x = 5 +/- root(6)/2
1039684
for x^2+5x-9=0  u can't do this the same way.9 * 1 = 9 but 9 + or - 1 not = 5 and 3 * 3 = 9 but 3 + or - 3 not = 5.\nso there is a formula u got to use here.\nit is  x = {- b + - sqrt(b^2 - 4ac)} / { 2a}\nso from the given quadratic equation,\na = 1,b = 5 and c = 9\nand then u substitute , u will get 2 values of x and that's the answer.remeber we have to use this formula if we are unable to factorise the way u did for the previous one.
941367
2x - 5x^1/2 - 3 = 0\n2x-3= 5x^1/2\nsquare both sides:\n(2x-3)^2 = 25x\n4x^2 - 12x + 9 = 25x\n4x^2 -37x +9 =0\nx= ( 37 +- sqrt( 37^2 -4(4)(9) ) /2(4)\nx= ( 37 +- sqrt( 1225 ) /2(4)\nx= ( 37 +- 35 )/2(4)\nx=  72/8 or x= 2/8\nx= 9 or x = 1/4
71189
x = 4\n\nSimplifying\n2x + -3 = 5\nReorder the terms:\n-3 + 2x = 5\nSolving\n-3 + 2x = 5\nSolving for variable 'x'.\n\nMove all terms containing x to the left, all other terms to the right.\n\nAdd '3' to each side of the equation.\n-3 + 3 + 2x = 5 + 3\nCombine like terms: -3 + 3 = 0\n0 + 2x = 5 + 3\n2x = 5 + 3\nCombine like terms: 5 + 3 = 8\n2x = 8\nDivide each side by '2'.\nx = 4\nSimplifying\nx = 4\n\nNow prove it.  (2*4)-3=5
952526
x^2-5x-2=0\n\nuse the quadratic formula\n\n[5  +/- sqrt(25-4*1*-2)]/2\n\n[5  +/-   sqrt(25+ 8)]/2\n\nI distributed the 1/2...easier to type\n\n5/2  +/-sqrt33/2
790653
In order to solve, you need a common denominator:\n2(sqrt(3/5)) - 5(sqrt (5/3))\n2(sqrt 3/sqrt 5) - 5 (sqrt 5/sqrt 3)\n2(sqrt3/sqrt3)(sqrt 3/sqrt 5) - 5(sqrt 5/sqrt 5)(sqrt 5/sqrt 3)\n2(3/sqrt 15) - 5(5/sqrt 15)\n6/sqrt 15 - 25/sqrt 15\n-19/sqrt 15
149693
3 - 2x = 5\n--> -2x = 5 - 3\n--> -2x = 2\n--> x = 2/(-2)\n--> x = -1 the answer!\n\nYou can check the answer by substituting the vale in the original eqn"\n3 - 2x\n= 3 - 2 (-1)\n= 3 + 2\n= 5
845970
25x² - 20x = 1\n25x² - 20x - 1 = 0\n25x² - 20x - 1 + 5 = 0 + 5\n25x² - 20x + 4 = 5\n(5x - 2)² = 5\n5x - 2 = +/- √5\n5x = 2 +/- √5\nx = (2 +/- √5) / 5\nx = (2 + √5) / 5 and (2 - √5) / 5
1058564
5x-3 = 3x +37\n2x=40\nx=20\n\nmuch better this time.  still kinda a dumb question.
431382
let the 2 numbers be x and y :\n\nx - y=23----A\nxy = -120 ---B\nx = -120/y\n\nsubstituting x in A:\n-120/y - y = 23 \n-120/y - y2/y = 23\n-120/y - 23 = y2/y\n-120/y - 23/y = y2/y \n-120-23/y = y\n-120 - 23 = y2\n0 = y2 + 23 + 120\nsolving the quadratic equation by the formula;\ny = -b (sq.root) b2-4ac/ 2a\n\ny= -8 and x is 15\nand then 15-(-8) = 23 \n         15x(-8) = -120
137368
2x+3 = 5\nsubtract 3 from both sides\n2x+3-3 = 5-3\n2x = 2\nDivide both sides by 2\nx=2/2\nx=1
1174116
8x^2 -5 -3x^2 + 7\n8x^2 - 3x^2 -5 + 7\n5x^2 + 2
607838
where is the rest of the equation?\n\n2x+5-x+3x+2-2x+5x-3x+8-4x+2+1x\n\nCollect like terms:\n\n2x-x+3x-2x+5x-3x-4x+1x+5+2+8+2\n\nRelocate the terms:\n\n1x+2x+3x+5x-x-2x-3x-4x+5+2+8+2 = 37\n\nAdd like terms:\n\n11x - 10x + 17 = 37\n\nNow, subtract all terms and solve from there.\n\nx + 17 = 37\nx = 20
839760
The roots are 0 and -16/5\n\nTo solve any quadratic equation eg. ax^2+bx+c, you can apply formula x= (-b+-sqrt(b^2-4ac))/2a