4x2-20x+10=0

Solution for 4x2-20x+10=0 equation:

4x^2-20x+10=0

a = 4; b = -20; c = +10;
Δ = b2-4ac
Δ = -202-4·4·10
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{15}}{2*4}=\frac{20-4\sqrt{15}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{15}}{2*4}=\frac{20+4\sqrt{15}}{8}
$