3x^2-8x=x^2-4x+48

Solution for 3x^2-8x=x^2-4x+48 equation:

3x^2-8x=x^2-4x+48

We move all terms to the left:

3x^2-8x-(x^2-4x+48)=0

We get rid of parentheses

3x^2-x^2-8x+4x-48=0

We add all the numbers together, and all the variables

2x^2-4x-48=0

a = 2; b = -4; c = -48;
Δ = b2-4ac
Δ = -42-4·2·(-48)
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{-(-4)-20}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*2}=\frac{24}{4}
=6
$