16x^2-52=6x^2+68

Solution for 16x^2-52=6x^2+68 equation:

16x^2-52=6x^2+68

We move all terms to the left:

16x^2-52-(6x^2+68)=0

We get rid of parentheses

16x^2-6x^2-68-52=0

We add all the numbers together, and all the variables

10x^2-120=0

a = 10; b = 0; c = -120;
Δ = b2-4ac
Δ = 02-4·10·(-120)
Δ = 4800
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:

$\sqrt{\Delta}=\sqrt{4800}=\sqrt{1600*3}=\sqrt{1600}*\sqrt{3}=40\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{3}}{2*10}=\frac{0-40\sqrt{3}}{20}
=-\frac{40\sqrt{3}}{20}
=-2\sqrt{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{3}}{2*10}=\frac{0+40\sqrt{3}}{20}
=\frac{40\sqrt{3}}{20}
=2\sqrt{3}
$