2x^2+48=3x^2+32x

Solution for 2x^2+48=3x^2+32x equation:

2x^2+48=3x^2+32x

We move all terms to the left:

2x^2+48-(3x^2+32x)=0

We get rid of parentheses

2x^2-3x^2-32x+48=0

We add all the numbers together, and all the variables

-1x^2-32x+48=0

a = -1; b = -32; c = +48;
Δ = b2-4ac
Δ = -322-4·(-1)·48
Δ = 1216
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:

$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{19}}{2*-1}=\frac{32-8\sqrt{19}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{19}}{2*-1}=\frac{32+8\sqrt{19}}{-2}
$