3x^2-40=x^2+116

Solution for 3x^2-40=x^2+116 equation:

3x^2-40=x^2+116

We move all terms to the left:

3x^2-40-(x^2+116)=0

We get rid of parentheses

3x^2-x^2-116-40=0

We add all the numbers together, and all the variables

2x^2-156=0

a = 2; b = 0; c = -156;
Δ = b2-4ac
Δ = 02-4·2·(-156)
Δ = 1248
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:

$\sqrt{\Delta}=\sqrt{1248}=\sqrt{16*78}=\sqrt{16}*\sqrt{78}=4\sqrt{78}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{78}}{2*2}=\frac{0-4\sqrt{78}}{4}
=-\frac{4\sqrt{78}}{4}
=-\sqrt{78}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{78}}{2*2}=\frac{0+4\sqrt{78}}{4}
=\frac{4\sqrt{78}}{4}
=\sqrt{78}
$