x^2-3x=2x^2+5x-48

Solution for x^2-3x=2x^2+5x-48 equation:

x^2-3x=2x^2+5x-48

We move all terms to the left:

x^2-3x-(2x^2+5x-48)=0

We get rid of parentheses

x^2-2x^2-3x-5x+48=0

We add all the numbers together, and all the variables

-1x^2-8x+48=0

a = -1; b = -8; c = +48;
Δ = b2-4ac
Δ = -82-4·(-1)·48
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*-1}=\frac{24}{-2}
=-12
$