2x+3x^2=6x+8x^2=

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

2x+3x^2=6x+8x^2=

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-5x^2-4x=0

a = -5; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-5)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-5}=\frac{0}{-10}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-5}=\frac{8}{-10}
=-4/5
$