(x+10)(2x+5)-(x+2)(x-2)=150

Solution for (x+10)(2x+5)-(x+2)(x-2)=150 equation:

(x+10)(2x+5)-(x+2)(x-2)=150

We move all terms to the left:

(x+10)(2x+5)-(x+2)(x-2)-(150)=0

We use the square of the difference formula

x^2+(x+10)(2x+5)+4-150=0

We multiply parentheses ..

x^2+(+2x^2+5x+20x+50)+4-150=0

We add all the numbers together, and all the variables

x^2+(+2x^2+5x+20x+50)-146=0

We get rid of parentheses

x^2+2x^2+5x+20x+50-146=0

We add all the numbers together, and all the variables

3x^2+25x-96=0

a = 3; b = 25; c = -96;
Δ = b2-4ac
Δ = 252-4·3·(-96)
Δ = 1777
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-\sqrt{1777}}{2*3}=\frac{-25-\sqrt{1777}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{1777}}{2*3}=\frac{-25+\sqrt{1777}}{6}
$