16x-2x^2+20=0

Solution for 16x-2x^2+20=0 equation:

16x-2x^2+20=0

a = -2; b = 16; c = +20;
Δ = b2-4ac
Δ = 162-4·(-2)·20
Δ = 416
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:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{26}}{2*-2}=\frac{-16-4\sqrt{26}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{26}}{2*-2}=\frac{-16+4\sqrt{26}}{-4}
$