3x^2-5x-28=4x^2+6x

Solution for 3x^2-5x-28=4x^2+6x equation:

3x^2-5x-28=4x^2+6x

We move all terms to the left:

3x^2-5x-28-(4x^2+6x)=0

We get rid of parentheses

3x^2-4x^2-5x-6x-28=0

We add all the numbers together, and all the variables

-1x^2-11x-28=0

a = -1; b = -11; c = -28;
Δ = b2-4ac
Δ = -112-4·(-1)·(-28)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3}{2*-1}=\frac{8}{-2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3}{2*-1}=\frac{14}{-2}
=-7
$