4x^2-20x+7=0

Solution for 4x^2-20x+7=0 equation:

4x^2-20x+7=0

a = 4; b = -20; c = +7;
Δ = b2-4ac
Δ = -202-4·4·7
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-12\sqrt{2}}{2*4}=\frac{20-12\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+12\sqrt{2}}{2*4}=\frac{20+12\sqrt{2}}{8}
$