(2x+10)(2x-5)=129

Solution for (2x+10)(2x-5)=129 equation:

(2x+10)(2x-5)=129

We move all terms to the left:

(2x+10)(2x-5)-(129)=0

We multiply parentheses ..

(+4x^2-10x+20x-50)-129=0

We get rid of parentheses

4x^2-10x+20x-50-129=0

We add all the numbers together, and all the variables

4x^2+10x-179=0

a = 4; b = 10; c = -179;
Δ = b2-4ac
Δ = 102-4·4·(-179)
Δ = 2964
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:

$\sqrt{\Delta}=\sqrt{2964}=\sqrt{4*741}=\sqrt{4}*\sqrt{741}=2\sqrt{741}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{741}}{2*4}=\frac{-10-2\sqrt{741}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{741}}{2*4}=\frac{-10+2\sqrt{741}}{8}
$