(10+2x)(6+2x)=140

Solution for (10+2x)(6+2x)=140 equation:

(10+2x)(6+2x)=140

We move all terms to the left:

(10+2x)(6+2x)-(140)=0

We add all the numbers together, and all the variables

(2x+10)(2x+6)-140=0

We multiply parentheses ..

(+4x^2+12x+20x+60)-140=0

We get rid of parentheses

4x^2+12x+20x+60-140=0

We add all the numbers together, and all the variables

4x^2+32x-80=0

a = 4; b = 32; c = -80;
Δ = b2-4ac
Δ = 322-4·4·(-80)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-48}{2*4}=\frac{-80}{8}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+48}{2*4}=\frac{16}{8}
=2
$