180=3x*(2x+5)+(x-5)

Solution for 180=3x*(2x+5)+(x-5) equation:

180=3x*(2x+5)+(x-5)

We move all terms to the left:

180-(3x*(2x+5)+(x-5))=0


We calculate terms in parentheses: -(3x*(2x+5)+(x-5)), so:

3x*(2x+5)+(x-5)

We multiply parentheses

6x^2+15x+(x-5)

We get rid of parentheses

6x^2+15x+x-5

We add all the numbers together, and all the variables

6x^2+16x-5

Back to the equation:

-(6x^2+16x-5)


We get rid of parentheses

-6x^2-16x+5+180=0

We add all the numbers together, and all the variables

-6x^2-16x+185=0

a = -6; b = -16; c = +185;
Δ = b2-4ac
Δ = -162-4·(-6)·185
Δ = 4696
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4696}=\sqrt{4*1174}=\sqrt{4}*\sqrt{1174}=2\sqrt{1174}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{1174}}{2*-6}=\frac{16-2\sqrt{1174}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{1174}}{2*-6}=\frac{16+2\sqrt{1174}}{-12}
$