(2x^2+7x-5)+(-8x^2-3x+5)=0

Solution for (2x^2+7x-5)+(-8x^2-3x+5)=0 equation:

(2x^2+7x-5)+(-8x^2-3x+5)=0

We get rid of parentheses

-8x^2+2x^2-3x+7x+5-5=0

We add all the numbers together, and all the variables

-6x^2+4x=0

a = -6; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-6)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-6}=\frac{-8}{-12}
=2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-6}=\frac{0}{-12}
=0
$