4x^2+28x+40=8

Solution for 4x^2+28x+40=8 equation:

4x^2+28x+40=8

We move all terms to the left:

4x^2+28x+40-(8)=0

We add all the numbers together, and all the variables

4x^2+28x+32=0

a = 4; b = 28; c = +32;
Δ = b2-4ac
Δ = 282-4·4·32
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{17}}{2*4}=\frac{-28-4\sqrt{17}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{17}}{2*4}=\frac{-28+4\sqrt{17}}{8}
$