2x^2+20x-300=-2x^2-20+500

Solution for 2x^2+20x-300=-2x^2-20+500 equation:

2x^2+20x-300=-2x^2-20+500

We move all terms to the left:

2x^2+20x-300-(-2x^2-20+500)=0

We get rid of parentheses

2x^2+2x^2+20x+20-500-300=0

We add all the numbers together, and all the variables

4x^2+20x-780=0

a = 4; b = 20; c = -780;
Δ = b2-4ac
Δ = 202-4·4·(-780)
Δ = 12880
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:

$\sqrt{\Delta}=\sqrt{12880}=\sqrt{16*805}=\sqrt{16}*\sqrt{805}=4\sqrt{805}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{805}}{2*4}=\frac{-20-4\sqrt{805}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{805}}{2*4}=\frac{-20+4\sqrt{805}}{8}
$