3x^2+56=x^2-24x

Solution for 3x^2+56=x^2-24x equation:

3x^2+56=x^2-24x

We move all terms to the left:

3x^2+56-(x^2-24x)=0

We get rid of parentheses

3x^2-x^2+24x+56=0

We add all the numbers together, and all the variables

2x^2+24x+56=0

a = 2; b = 24; c = +56;
Δ = b2-4ac
Δ = 242-4·2·56
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{2}}{2*2}=\frac{-24-8\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{2}}{2*2}=\frac{-24+8\sqrt{2}}{4}
$