(2x^2+5x-8)=(3x^2-6x+2)

Solution for (2x^2+5x-8)=(3x^2-6x+2) equation:

(2x^2+5x-8)=(3x^2-6x+2)

We move all terms to the left:

(2x^2+5x-8)-((3x^2-6x+2))=0

We get rid of parentheses

2x^2+5x-((3x^2-6x+2))-8=0


We calculate terms in parentheses: -((3x^2-6x+2)), so:

(3x^2-6x+2)

We get rid of parentheses

3x^2-6x+2

Back to the equation:

-(3x^2-6x+2)


We get rid of parentheses

2x^2-3x^2+5x+6x-2-8=0

We add all the numbers together, and all the variables

-1x^2+11x-10=0

a = -1; b = 11; c = -10;
Δ = b2-4ac
Δ = 112-4·(-1)·(-10)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-9}{2*-1}=\frac{-20}{-2}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+9}{2*-1}=\frac{-2}{-2}
=1
$