2x^2+15=4x^2-35

Solution for 2x^2+15=4x^2-35 equation:

2x^2+15=4x^2-35

We move all terms to the left:

2x^2+15-(4x^2-35)=0

We get rid of parentheses

2x^2-4x^2+35+15=0

We add all the numbers together, and all the variables

-2x^2+50=0

a = -2; b = 0; c = +50;
Δ = b2-4ac
Δ = 02-4·(-2)·50
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*-2}=\frac{-20}{-4}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*-2}=\frac{20}{-4}
=-5
$