4x^2+42x=2x-8

Solution for 4x^2+42x=2x-8 equation:

4x^2+42x=2x-8

We move all terms to the left:

4x^2+42x-(2x-8)=0

We get rid of parentheses

4x^2+42x-2x+8=0

We add all the numbers together, and all the variables

4x^2+40x+8=0

a = 4; b = 40; c = +8;
Δ = b2-4ac
Δ = 402-4·4·8
Δ = 1472
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:

$\sqrt{\Delta}=\sqrt{1472}=\sqrt{64*23}=\sqrt{64}*\sqrt{23}=8\sqrt{23}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{23}}{2*4}=\frac{-40-8\sqrt{23}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{23}}{2*4}=\frac{-40+8\sqrt{23}}{8}
$