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-16x^{2}+32x+40=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{-32±\sqrt{32^{2}-4\left(-16\right)\times 40}}{2\left(-16\right)}
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{-32±\sqrt{1024-4\left(-16\right)\times 40}}{2\left(-16\right)}
Square 32.
x=\frac{-32±\sqrt{1024+64\times 40}}{2\left(-16\right)}
Multiply -4 times -16.
x=\frac{-32±\sqrt{1024+2560}}{2\left(-16\right)}
Multiply 64 times 40.
x=\frac{-32±\sqrt{3584}}{2\left(-16\right)}
Add 1024 to 2560.
x=\frac{-32±16\sqrt{14}}{2\left(-16\right)}
Take the square root of 3584.
x=\frac{-32±16\sqrt{14}}{-32}
Multiply 2 times -16.
x=\frac{16\sqrt{14}-32}{-32}
Now solve the equation x=\frac{-32±16\sqrt{14}}{-32} when ± is plus. Add -32 to 16\sqrt{14}.
x=-\frac{\sqrt{14}}{2}+1
Divide -32+16\sqrt{14} by -32.
x=\frac{-16\sqrt{14}-32}{-32}
Now solve the equation x=\frac{-32±16\sqrt{14}}{-32} when ± is minus. Subtract 16\sqrt{14} from -32.
x=\frac{\sqrt{14}}{2}+1
Divide -32-16\sqrt{14} by -32.
-16x^{2}+32x+40=-16\left(x-\left(-\frac{\sqrt{14}}{2}+1\right)\right)\left(x-\left(\frac{\sqrt{14}}{2}+1\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1-\frac{\sqrt{14}}{2} for x_{1} and 1+\frac{\sqrt{14}}{2} for x_{2}.
x ^ 2 -2x -\frac{5}{2} = 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.
r + s = 2 rs = -\frac{5}{2}
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 = 1 - u s = 1 + u
Two numbers r and s sum up to 2 exactly when the average of the two numbers is \frac{1}{2}*2 = 1. 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>
(1 - u) (1 + u) = -\frac{5}{2}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{2}
1 - u^2 = -\frac{5}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{5}{2}-1 = -\frac{7}{2}
Simplify the expression by subtracting 1 on both sides
u^2 = \frac{7}{2} u = \pm\sqrt{\frac{7}{2}} = \pm \frac{\sqrt{7}}{\sqrt{2}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =1 - \frac{\sqrt{7}}{\sqrt{2}} = -0.871 s = 1 + \frac{\sqrt{7}}{\sqrt{2}} = 2.871
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.