3x^2-8x-28=2x^2+3x

Solution for 3x^2-8x-28=2x^2+3x equation:

3x^2-8x-28=2x^2+3x

We move all terms to the left:

3x^2-8x-28-(2x^2+3x)=0

We get rid of parentheses

3x^2-2x^2-8x-3x-28=0

We add all the numbers together, and all the variables

x^2-11x-28=0

a = 1; b = -11; c = -28;
Δ = b2-4ac
Δ = -112-4·1·(-28)
Δ = 233
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{233}}{2*1}=\frac{11-\sqrt{233}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{233}}{2*1}=\frac{11+\sqrt{233}}{2}
$