3x(2x+20)+(4x-20)=x

Solution for 3x(2x+20)+(4x-20)=x equation:

3x(2x+20)+(4x-20)=x

We move all terms to the left:

3x(2x+20)+(4x-20)-(x)=0

We add all the numbers together, and all the variables

-1x+3x(2x+20)+(4x-20)=0

We multiply parentheses

6x^2-1x+60x+(4x-20)=0

We get rid of parentheses

6x^2-1x+60x+4x-20=0

We add all the numbers together, and all the variables

6x^2+63x-20=0

a = 6; b = 63; c = -20;
Δ = b2-4ac
Δ = 632-4·6·(-20)
Δ = 4449
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(63)-\sqrt{4449}}{2*6}=\frac{-63-\sqrt{4449}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+\sqrt{4449}}{2*6}=\frac{-63+\sqrt{4449}}{12}
$