20+(6x^2)=20+x

Solution for 20+(6x^2)=20+x equation:

20+(6x^2)=20+x

We move all terms to the left:

20+(6x^2)-(20+x)=0

We add all the numbers together, and all the variables

6x^2-(x+20)+20=0

We get rid of parentheses

6x^2-x-20+20=0

We add all the numbers together, and all the variables

6x^2-1x=0

a = 6; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·6·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*6}=\frac{0}{12}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*6}=\frac{2}{12}
=1/6
$