3x^2+13x-(28)=2x^2+(2)

Solution for 3x^2+13x-(28)=2x^2+(2) equation:

3x^2+13x-(28)=2x^2+(2)

We move all terms to the left:

3x^2+13x-(28)-(2x^2+(2))=0

We get rid of parentheses

3x^2-2x^2+13x-2-28=0

We add all the numbers together, and all the variables

x^2+13x-30=0

a = 1; b = 13; c = -30;
Δ = b2-4ac
Δ = 132-4·1·(-30)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*1}=\frac{-30}{2}
=-15
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*1}=\frac{4}{2}
=2
$