sove this quadratic equation?
819448
2p^3 = 2p(p + 2)\n2p^3-2p(p+2)=0\n2p(p^2-p-2)=0\n2p(p-2)(p+1)\np=0\np-2=0\np=2\np+1=0\np=-1\n\nroots are \np=0, 2, -1
412542
x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
605879
(x+3)(x+12)(x+2)(x+8)= [(x+3)(x+8)][(x+12)(x+2)]\n                     = (x^2+11x+24)*(x^2+14x+24)\nlet t=x^2+24+(25X)&#92;2 . \nSo we have: x^2+11x+24=t-(3x)&#92;2 and x^2+14x+24=t+(3x)&#92;2                                 \nso the equation becomes: [t-(3x)&#92;2][t+(3x)&#92;2] = 4x^2\n                    <-->t^2-(9x^2)&#92;4=4x^2  \n                    <-->t^2=(25x^2)&#92;4\n                    <-->t=(5x)&#92;2 or t=-(5x)&#92;2\n                    <-->x^2+10x+24=0 or x^2+15x+24=0\ncontinue by yourself
123389
distribute the x to all terms inside the equation\n\n2x^2-26x
983225
remember the rule nx^n-1\n2x^2-1\n \nSo in this case it is 2x
538847
For any quadratic of the form ax^2 + bx + c = 0, the roots are\n\nx = (-b +/- sqrt(b^2-4ac)) / 2a\n\nSqrt here is shorthand for "the square root of".
896553
4x^2+8x-x-2\n(4x-1)(x+2)
316686
the process u follow is correct, but actually it is wrong. fundamentally wat u did is ok, but technically it is wrong. because this becomes a quadratic equation which has more than 1 root. so this equation has more than 1 answers for which it is true.
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
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
175133
2x-1(xsquared-x+10)squared
848564
Substitute 2x-1 for y in the middle part, you get:\nx^2 + x -6(2x-1) = 2x-1\nThis you can reduce to a quadratic equation for x.
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
464039
Your equation would be:\n\n= ((1/p) - (1/q))/(1/pq)\n\n= ((q - p)/pq) * pq\n\n= q - p\n\nHope I am clear.
893823
Use the quadratic equation \n\nt = -b + or - sqr(b^2 - 4ac)/2a\n\nyour equation is already in the form:\n\nat^2 + bt + c = 0\n\njust plug in the variables and solve for t twice
670824
There are 2 ways to solve for this \n\n1. Separating the x^2\n\n2. Quadratic Formula\n   ax^2 + bx + c = 0\n x = (-b + or - sqrt(b^2 - 4ac))/2a\n http://www.1728.com/quadrtc.gif\n\n\nSolving it by separating x^2\n\na. First separate x^2\n(x + ?)(x +?)\n\nb. Now we have to figure out the 2 variables that would give us 4x and -12 when we multiply them out. notice, that 4 is positive so the larger number has to be Positive.\n\n-12 = 12 * -1   12x + -1x = 11x\n-12 = 6 * -2    6x + -2x = 4x\n-12 = 4 * -3    4x + -3x = x\n\ntherefore the 2 variables would be 6 and -2\n\n(x + 6)(x - 2) = 0\nSo x, would be either -6, or 2\n---------------------------------------------  \n2. SOLVING USING THE QUADRATIC FORMULA\nThis one is way easier\n\nax^2 + bx + c = 0\nx^2 + 4x - 12 = 0\n\n x = (-b + or - sqrt(b^2 - 4ac))/2a\n x = (-4 + or - sqrt(4^2 - 4*1*-12))/2*1\n\n x = (-4 + or - 8)/2\n x = (-4 + 8)/2 = 2\n\n x - (-4 - 8)/2 = -6
925669
You need to use the quadratic equation...\n\nhttp://mathworld.wolfram.com/QuadraticEquation.html\n\nOR\n\nx (x - 5) = 0