(3x^2+4x-8)=(-2x^2+4x+2)=

Solution for (3x^2+4x-8)=(-2x^2+4x+2)= equation:

(3x^2+4x-8)=(-2x^2+4x+2)=

We move all terms to the left:

(3x^2+4x-8)-((-2x^2+4x+2))=0

We get rid of parentheses

-((-2x^2+4x+2))+3x^2+4x-8=0


We calculate terms in parentheses: -((-2x^2+4x+2)), so:

(-2x^2+4x+2)

We get rid of parentheses

-2x^2+4x+2

Back to the equation:

-(-2x^2+4x+2)


We add all the numbers together, and all the variables

3x^2-(-2x^2+4x+2)+4x-8=0

We get rid of parentheses

3x^2+2x^2-4x+4x-2-8=0

We add all the numbers together, and all the variables

5x^2-10=0

a = 5; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·5·(-10)
Δ = 200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{2}}{2*5}=\frac{0-10\sqrt{2}}{10}
=-\frac{10\sqrt{2}}{10}
=-\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{2}}{2*5}=\frac{0+10\sqrt{2}}{10}
=\frac{10\sqrt{2}}{10}
=\sqrt{2}
$