(3x2-28)=(2x2-3x)

Solution for (3x2-28)=(2x2-3x) equation:

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

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -((2x^2-3x)), so:

(2x^2-3x)

We get rid of parentheses

2x^2-3x

Back to the equation:

-(2x^2-3x)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+3x-28=0

a = 1; b = 3; c = -28;
Δ = b2-4ac
Δ = 32-4·1·(-28)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*1}=\frac{-14}{2}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*1}=\frac{8}{2}
=4
$