(3+2x)(5+2x)=20

Solution for (3+2x)(5+2x)=20 equation:

(3+2x)(5+2x)=20

We move all terms to the left:

(3+2x)(5+2x)-(20)=0

We add all the numbers together, and all the variables

(2x+3)(2x+5)-20=0

We multiply parentheses ..

(+4x^2+10x+6x+15)-20=0

We get rid of parentheses

4x^2+10x+6x+15-20=0

We add all the numbers together, and all the variables

4x^2+16x-5=0

a = 4; b = 16; c = -5;
Δ = b2-4ac
Δ = 162-4·4·(-5)
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{21}}{2*4}=\frac{-16-4\sqrt{21}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{21}}{2*4}=\frac{-16+4\sqrt{21}}{8}
$