5x^2-20x+16=0

Solution for 5x^2-20x+16=0 equation:

5x^2-20x+16=0

a = 5; b = -20; c = +16;
Δ = b2-4ac
Δ = -202-4·5·16
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{5}}{2*5}=\frac{20-4\sqrt{5}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{5}}{2*5}=\frac{20+4\sqrt{5}}{10}
$