3x^2+8x=6x+13x2+8x=6x+1

Solution for 3x^2+8x=6x+13x2+8x=6x+1 equation:

3x^2+8x=6x+13x^2+8x=6x+1

We move all terms to the left:

3x^2+8x-(6x+13x^2+8x)=0

We get rid of parentheses

3x^2-13x^2-6x-8x+8x=0

We add all the numbers together, and all the variables

-10x^2-6x=0

a = -10; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·(-10)·0
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{-(-6)-6}{2*-10}=\frac{0}{-20}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*-10}=\frac{12}{-20}
=-3/5
$