Simplify by using the quadratic formula: 3X^2-5X+2=0?
139621
Ran into a little difficulty, I can't find my math editor.\n\nBut here goes:\nx = (-b +,- SQRT(b^2 - 4ac))/2a\na = 3, b = -5, c = 2;\nx = (-(-5) +,- SQRT(25 - 24))/ 6\nx = (5 + 1) / 6; x = (5 - 1) / 6\nx = (1, 2/3)\nThis would have looked a lot better if I had my math editor!!!\n\nClarification: I solved your quadratic equation using the quadratic formula.\n\nMashi below simplified your quadratic equation, which you don't need the quadratic formula to do. Substitute the values of x I obtained by solving into the simplified equation, they will equal 0
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
968026
I got   3(x² - 5x + 4) or 3( x - 4 )( x - 1 )\n\nLet me show you my work...\n\n3( x - 2 )² - 3(x - 2) - 6\n\nSimplify\n\n3( x -2 )( x - 2 ) - 3( x- 2 ) - 6\n\nDistribute\n\n3( x² - 4x + 4) - 3x + 6 - 6\n\nDistribute and Simplify\n\n3x² - 12x + 12 - 3x\n\nSimplify\n\n3x² - 15x + 12\n\nYou can now simplify by factoring out 3\n\n3(x² - 5x + 4)\nor\n3( x - 4 )( x - 1 )
734679
5a^2 + 3a - 2 = 0\n\ncan be factored as (5a - 2) (a + 1) = 0\nThus a = 2/5 and a = -1
364629
I am sorry that I can't explain how, but the steps and the answer is here... (6x)(5x+2)(3x-1)\n\n90x^3+6x^2-12x \n6 (15x^3+x^2-2x)\n6x (15x^2+x-2)\n(6x) (5x+2)(3x-1)\n\nTo check 5x*3x=15x^2, 2*3x=6x, 5x*-1=-5x,2*-1=-2, so\n15x^2+6x-5x-2, take out 5x out of 6x and you get 1x.\n\nThen multiply 6x across the whole formula.
394036
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
11476
algebraic simplification\n\nx over 4+3 = 2x+3 over 5\nx / (4+3) = (2x+3) / 5\nx/7 = (2x+3)/5\n5x/7 = 2x+3\n9x/7 = -3\n\nx = -7/3
665603
you put (x+2) in for x in the equation first\n\n((x+2)^3)+(2(x+2)^2)-4\n\nthen simplify \n\nx^3 + 8x^2 + 20x + 12\n\nand multiply the entire thing by 3\n\n3(x^3 + 8x^2 + 20x + 12)\n\n3x^3 + 24x^2 + 60x + 36
631123
x(5x-3)=1\nx(5x) + x (-3) = 1\n5x^2 -3x = 1\n5x^2 -3x -1 = 0
977979
x= 3/5\n\nIf you factor by grouping you would get\n25x^2 - 15x -15x + 9 = 0\n\nthen you would factor the first two terms and the last two terms\n5x(5x-3)   -3(5x-3)    =  0\n\nuse the terms outside the parentheses are one factor and the one inside the parentheses as the second factor\n(5x-3)(5x-3) = 0\n\nset to zero\n5x-3 = 0\n\ndo algebra\nx=3/5\n\nor you could have just noticed that that equation is a perfect square
695355
-5 -2(x+3) + 7x\n= -5 -2x -6 + 7x\n= -11 -2x + 7x\n= 5x - 11
866633
9x^2+18x+8=0\n\n(3x + 2)(3x + 4)=0\n3x+2=0\nx=-2/3\n3x+4=0\nx=-4/3
925669
You need to use the quadratic equation...\n\nhttp://mathworld.wolfram.com/QuadraticEquation.html\n\nOR\n\nx (x - 5) = 0
551906
Add each equation\nA + B = 3x^2 + 2x - 2\nA - B = -x^2 + 4x - 8\n2A = 2x^2 + 6x - 10\nDivide each side by 2\nA = x^2 + 3x - 5\nNow plug into one of the original equations to get B\nB = 3x^2 + 2x - 2 - (x^2 + 3x - 5)\nB = 2x^2 - x + 3
764068
2x - 2 = x^2 - 4x + 3\nSet one side equal to 0\n0 = x^2 - 6x + 5\nFactor\n0 = (x - 5)(x - 1)\nSet each factor equal to 0\n0 = x - 5              0 = x - 1\nSolve\n5 = x                  1 = x
534396
quadratic equations of the form \n\nax^2+bx+c=0\n\ncan be used with the quadratic formula, see\n\nhttp://mathworld.wolfram.com/QuadraticFormula.html\n\nso just transform all your equations into the form\n\nax^2+bx+c = 0 (x can be called differently, of course).\n\nEdit: for example, let's take \n\n5t^2 + 20t - 60 = 0\n\naccording to the quadratic formula, the solutions are\n\nt_1 = (-20+sqrt(20*20-4*5*(-60))) / (2*5) = (-20+sqrt(1600))/10 = 2\n\nand\n\nt_2 = (-20-sqrt(20*20-4*5*(-60))) / (2*5) = (-20-sqrt(1600))/10 = -6\n\nNote: if the term below the root (sqrt) is negative, it is not a valid solution - sometimes the quadratic equation has only one solution, or none.
785524
y = 3x-5\n\nx^2 +(3x-5)^2 = 25\nx^2 + 9x^2 -30x + 25 = 25\n10x^2 -30x = 0\n10x(x-3) = 0\n\nso x = 0 or x = 3\n\nWhen x = 0 y = -5\n\nWhen x = 3  y = 4