2(2x^2+40x-525)=0

Solution for 2(2x^2+40x-525)=0 equation:

2(2x^2+40x-525)=0

We multiply parentheses

4x^2+80x-1050=0

a = 4; b = 80; c = -1050;
Δ = b2-4ac
Δ = 802-4·4·(-1050)
Δ = 23200
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:

$\sqrt{\Delta}=\sqrt{23200}=\sqrt{400*58}=\sqrt{400}*\sqrt{58}=20\sqrt{58}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-20\sqrt{58}}{2*4}=\frac{-80-20\sqrt{58}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+20\sqrt{58}}{2*4}=\frac{-80+20\sqrt{58}}{8}
$