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32x^{2}+12x-7=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-12±\sqrt{12^{2}-4\times 32\left(-7\right)}}{2\times 32}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-12±\sqrt{144-4\times 32\left(-7\right)}}{2\times 32}
Square 12.
x=\frac{-12±\sqrt{144-128\left(-7\right)}}{2\times 32}
Multiply -4 times 32.
x=\frac{-12±\sqrt{144+896}}{2\times 32}
Multiply -128 times -7.
x=\frac{-12±\sqrt{1040}}{2\times 32}
Add 144 to 896.
x=\frac{-12±4\sqrt{65}}{2\times 32}
Take the square root of 1040.
x=\frac{-12±4\sqrt{65}}{64}
Multiply 2 times 32.
x=\frac{4\sqrt{65}-12}{64}
Now solve the equation x=\frac{-12±4\sqrt{65}}{64} when ± is plus. Add -12 to 4\sqrt{65}.
x=\frac{\sqrt{65}-3}{16}
Divide -12+4\sqrt{65} by 64.
x=\frac{-4\sqrt{65}-12}{64}
Now solve the equation x=\frac{-12±4\sqrt{65}}{64} when ± is minus. Subtract 4\sqrt{65} from -12.
x=\frac{-\sqrt{65}-3}{16}
Divide -12-4\sqrt{65} by 64.
32x^{2}+12x-7=32\left(x-\frac{\sqrt{65}-3}{16}\right)\left(x-\frac{-\sqrt{65}-3}{16}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{65}}{16} for x_{1} and \frac{-3-\sqrt{65}}{16} for x_{2}.
x ^ 2 +\frac{3}{8}x -\frac{7}{32} = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 32
r + s = -\frac{3}{8} rs = -\frac{7}{32}
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = -\frac{3}{16} - u s = -\frac{3}{16} + u
Two numbers r and s sum up to -\frac{3}{8} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{8} = -\frac{3}{16}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-\frac{3}{16} - u) (-\frac{3}{16} + u) = -\frac{7}{32}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{7}{32}
\frac{9}{256} - u^2 = -\frac{7}{32}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{7}{32}-\frac{9}{256} = -\frac{65}{256}
Simplify the expression by subtracting \frac{9}{256} on both sides
u^2 = \frac{65}{256} u = \pm\sqrt{\frac{65}{256}} = \pm \frac{\sqrt{65}}{16}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{3}{16} - \frac{\sqrt{65}}{16} = -0.691 s = -\frac{3}{16} + \frac{\sqrt{65}}{16} = 0.316
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.