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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -((-6x+5)), so:

(-6x+5)

We get rid of parentheses

-6x+5

Back to the equation:

-(-6x+5)


We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2+11x-13=0

a = 2; b = 11; c = -13;
Δ = b2-4ac
Δ = 112-4·2·(-13)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-15}{2*2}=\frac{-26}{4}
=-6+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+15}{2*2}=\frac{4}{4}
=1
$