3x^2+9x-8=2x^2+6x-4

Solution for 3x^2+9x-8=2x^2+6x-4 equation:

3x^2+9x-8=2x^2+6x-4

We move all terms to the left:

3x^2+9x-8-(2x^2+6x-4)=0

We get rid of parentheses

3x^2-2x^2+9x-6x+4-8=0

We add all the numbers together, and all the variables

x^2+3x-4=0

a = 1; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·1·(-4)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*1}=\frac{2}{2}
=1
$