4x-20x^2+64=0

Solution for 4x-20x^2+64=0 equation:

4x-20x^2+64=0

a = -20; b = 4; c = +64;
Δ = b2-4ac
Δ = 42-4·(-20)·64
Δ = 5136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5136}=\sqrt{16*321}=\sqrt{16}*\sqrt{321}=4\sqrt{321}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{321}}{2*-20}=\frac{-4-4\sqrt{321}}{-40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{321}}{2*-20}=\frac{-4+4\sqrt{321}}{-40}
$