(10+2x)(15+2x)-150=150

Solution for (10+2x)(15+2x)-150=150 equation:

(10+2x)(15+2x)-150=150

We move all terms to the left:

(10+2x)(15+2x)-150-(150)=0

We add all the numbers together, and all the variables

(2x+10)(2x+15)-150-150=0

We add all the numbers together, and all the variables

(2x+10)(2x+15)-300=0

We multiply parentheses ..

(+4x^2+30x+20x+150)-300=0

We get rid of parentheses

4x^2+30x+20x+150-300=0

We add all the numbers together, and all the variables

4x^2+50x-150=0

a = 4; b = 50; c = -150;
Δ = b2-4ac
Δ = 502-4·4·(-150)
Δ = 4900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4900}=70$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-70}{2*4}=\frac{-120}{8}
=-15
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+70}{2*4}=\frac{20}{8}
=2+1/2
$