(2x+5)(2x+2)=168

Solution for (2x+5)(2x+2)=168 equation:

(2x+5)(2x+2)=168

We move all terms to the left:

(2x+5)(2x+2)-(168)=0

We multiply parentheses ..

(+4x^2+4x+10x+10)-168=0

We get rid of parentheses

4x^2+4x+10x+10-168=0

We add all the numbers together, and all the variables

4x^2+14x-158=0

a = 4; b = 14; c = -158;
Δ = b2-4ac
Δ = 142-4·4·(-158)
Δ = 2724
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:

$\sqrt{\Delta}=\sqrt{2724}=\sqrt{4*681}=\sqrt{4}*\sqrt{681}=2\sqrt{681}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{681}}{2*4}=\frac{-14-2\sqrt{681}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{681}}{2*4}=\frac{-14+2\sqrt{681}}{8}
$