4x^2+20x+25=8

Solution for 4x^2+20x+25=8 equation:

4x^2+20x+25=8

We move all terms to the left:

4x^2+20x+25-(8)=0

We add all the numbers together, and all the variables

4x^2+20x+17=0

a = 4; b = 20; c = +17;
Δ = b2-4ac
Δ = 202-4·4·17
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{2}}{2*4}=\frac{-20-8\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{2}}{2*4}=\frac{-20+8\sqrt{2}}{8}
$