(4x^2+2x-1)=(3x^2+2x+8)

Solution for (4x^2+2x-1)=(3x^2+2x+8) equation:

(4x^2+2x-1)=(3x^2+2x+8)

We move all terms to the left:

(4x^2+2x-1)-((3x^2+2x+8))=0

We get rid of parentheses

4x^2+2x-((3x^2+2x+8))-1=0


We calculate terms in parentheses: -((3x^2+2x+8)), so:

(3x^2+2x+8)

We get rid of parentheses

3x^2+2x+8

Back to the equation:

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


We get rid of parentheses

4x^2-3x^2+2x-2x-8-1=0

We add all the numbers together, and all the variables

x^2-9=0

a = 1; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·1·(-9)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*1}=\frac{6}{2}
=3
$