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32x^{2}-5x-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{-\left(-5\right)±\sqrt{\left(-5\right)^{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{-\left(-5\right)±\sqrt{25-4\times 32\left(-7\right)}}{2\times 32}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-128\left(-7\right)}}{2\times 32}
Multiply -4 times 32.
x=\frac{-\left(-5\right)±\sqrt{25+896}}{2\times 32}
Multiply -128 times -7.
x=\frac{-\left(-5\right)±\sqrt{921}}{2\times 32}
Add 25 to 896.
x=\frac{5±\sqrt{921}}{2\times 32}
The opposite of -5 is 5.
x=\frac{5±\sqrt{921}}{64}
Multiply 2 times 32.
x=\frac{\sqrt{921}+5}{64}
Now solve the equation x=\frac{5±\sqrt{921}}{64} when ± is plus. Add 5 to \sqrt{921}.
x=\frac{5-\sqrt{921}}{64}
Now solve the equation x=\frac{5±\sqrt{921}}{64} when ± is minus. Subtract \sqrt{921} from 5.
32x^{2}-5x-7=32\left(x-\frac{\sqrt{921}+5}{64}\right)\left(x-\frac{5-\sqrt{921}}{64}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5+\sqrt{921}}{64} for x_{1} and \frac{5-\sqrt{921}}{64} for x_{2}.
x ^ 2 -\frac{5}{32}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{5}{32} 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{5}{64} - u s = \frac{5}{64} + u
Two numbers r and s sum up to \frac{5}{32} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{32} = \frac{5}{64}. 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{5}{64} - u) (\frac{5}{64} + u) = -\frac{7}{32}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{7}{32}
\frac{25}{4096} - u^2 = -\frac{7}{32}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{7}{32}-\frac{25}{4096} = -\frac{921}{4096}
Simplify the expression by subtracting \frac{25}{4096} on both sides
u^2 = \frac{921}{4096} u = \pm\sqrt{\frac{921}{4096}} = \pm \frac{\sqrt{921}}{64}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{64} - \frac{\sqrt{921}}{64} = -0.396 s = \frac{5}{64} + \frac{\sqrt{921}}{64} = 0.552
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.