(4x-10)(2x+52)=132

Solution for (4x-10)(2x+52)=132 equation:

(4x-10)(2x+52)=132

We move all terms to the left:

(4x-10)(2x+52)-(132)=0

We multiply parentheses ..

(+8x^2+208x-20x-520)-132=0

We get rid of parentheses

8x^2+208x-20x-520-132=0

We add all the numbers together, and all the variables

8x^2+188x-652=0

a = 8; b = 188; c = -652;
Δ = b2-4ac
Δ = 1882-4·8·(-652)
Δ = 56208
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:

$\sqrt{\Delta}=\sqrt{56208}=\sqrt{16*3513}=\sqrt{16}*\sqrt{3513}=4\sqrt{3513}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(188)-4\sqrt{3513}}{2*8}=\frac{-188-4\sqrt{3513}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(188)+4\sqrt{3513}}{2*8}=\frac{-188+4\sqrt{3513}}{16}
$