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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

(2x+10)(2x+6)-120=0

We multiply parentheses ..

(+4x^2+12x+20x+60)-120=0

We get rid of parentheses

4x^2+12x+20x+60-120=0

We add all the numbers together, and all the variables

4x^2+32x-60=0

a = 4; b = 32; c = -60;
Δ = b2-4ac
Δ = 322-4·4·(-60)
Δ = 1984
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:

$\sqrt{\Delta}=\sqrt{1984}=\sqrt{64*31}=\sqrt{64}*\sqrt{31}=8\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{31}}{2*4}=\frac{-32-8\sqrt{31}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{31}}{2*4}=\frac{-32+8\sqrt{31}}{8}
$