3x^2+84=4x^2+12

Solution for 3x^2+84=4x^2+12 equation:

3x^2+84=4x^2+12

We move all terms to the left:

3x^2+84-(4x^2+12)=0

We get rid of parentheses

3x^2-4x^2-12+84=0

We add all the numbers together, and all the variables

-1x^2+72=0

a = -1; b = 0; c = +72;
Δ = b2-4ac
Δ = 02-4·(-1)·72
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*-1}=\frac{0-12\sqrt{2}}{-2}
=-\frac{12\sqrt{2}}{-2}
=-\frac{6\sqrt{2}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*-1}=\frac{0+12\sqrt{2}}{-2}
=\frac{12\sqrt{2}}{-2}
=\frac{6\sqrt{2}}{-1}
$