8x^2+x-20=2x^2+8x

Solution for 8x^2+x-20=2x^2+8x equation:

8x^2+x-20=2x^2+8x

We move all terms to the left:

8x^2+x-20-(2x^2+8x)=0

We get rid of parentheses

8x^2-2x^2+x-8x-20=0

We add all the numbers together, and all the variables

6x^2-7x-20=0

a = 6; b = -7; c = -20;
Δ = b2-4ac
Δ = -72-4·6·(-20)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-23}{2*6}=\frac{-16}{12}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+23}{2*6}=\frac{30}{12}
=2+1/2
$