(5+5x)*(2+2x)=38

Solution for (5+5x)*(2+2x)=38 equation:

(5+5x)(2+2x)=38

We move all terms to the left:

(5+5x)(2+2x)-(38)=0

We add all the numbers together, and all the variables

(5x+5)(2x+2)-38=0

We multiply parentheses ..

(+10x^2+10x+10x+10)-38=0

We get rid of parentheses

10x^2+10x+10x+10-38=0

We add all the numbers together, and all the variables

10x^2+20x-28=0

a = 10; b = 20; c = -28;
Δ = b2-4ac
Δ = 202-4·10·(-28)
Δ = 1520
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:

$\sqrt{\Delta}=\sqrt{1520}=\sqrt{16*95}=\sqrt{16}*\sqrt{95}=4\sqrt{95}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{95}}{2*10}=\frac{-20-4\sqrt{95}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{95}}{2*10}=\frac{-20+4\sqrt{95}}{20}
$