(2x-29)(10x+5)=180

Solution for (2x-29)(10x+5)=180 equation:

(2x-29)(10x+5)=180

We move all terms to the left:

(2x-29)(10x+5)-(180)=0

We multiply parentheses ..

(+20x^2+10x-290x-145)-180=0

We get rid of parentheses

20x^2+10x-290x-145-180=0

We add all the numbers together, and all the variables

20x^2-280x-325=0

a = 20; b = -280; c = -325;
Δ = b2-4ac
Δ = -2802-4·20·(-325)
Δ = 104400
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:

$\sqrt{\Delta}=\sqrt{104400}=\sqrt{3600*29}=\sqrt{3600}*\sqrt{29}=60\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-280)-60\sqrt{29}}{2*20}=\frac{280-60\sqrt{29}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-280)+60\sqrt{29}}{2*20}=\frac{280+60\sqrt{29}}{40}
$