180=3x(6x+10)+x

Solution for 180=3x(6x+10)+x equation:

180=3x(6x+10)+x

We move all terms to the left:

180-(3x(6x+10)+x)=0


We calculate terms in parentheses: -(3x(6x+10)+x), so:

3x(6x+10)+x

We add all the numbers together, and all the variables

x+3x(6x+10)

We multiply parentheses

18x^2+x+30x

We add all the numbers together, and all the variables

18x^2+31x

Back to the equation:

-(18x^2+31x)


We get rid of parentheses

-18x^2-31x+180=0

a = -18; b = -31; c = +180;
Δ = b2-4ac
Δ = -312-4·(-18)·180
Δ = 13921
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{-(-31)-\sqrt{13921}}{2*-18}=\frac{31-\sqrt{13921}}{-36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{13921}}{2*-18}=\frac{31+\sqrt{13921}}{-36}
$