(10+2x)(6+2x)-60=36

Solution for (10+2x)(6+2x)-60=36 equation:

(10+2x)(6+2x)-60=36

We move all terms to the left:

(10+2x)(6+2x)-60-(36)=0

We add all the numbers together, and all the variables

(2x+10)(2x+6)-60-36=0

We add all the numbers together, and all the variables

(2x+10)(2x+6)-96=0

We multiply parentheses ..

(+4x^2+12x+20x+60)-96=0

We get rid of parentheses

4x^2+12x+20x+60-96=0

We add all the numbers together, and all the variables

4x^2+32x-36=0

a = 4; b = 32; c = -36;
Δ = b2-4ac
Δ = 322-4·4·(-36)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-40}{2*4}=\frac{-72}{8}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+40}{2*4}=\frac{8}{8}
=1
$