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

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

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

We move all terms to the left:

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

We multiply parentheses

6x^2+10x+(x-5)-180=0

We get rid of parentheses

6x^2+10x+x-5-180=0

We add all the numbers together, and all the variables

6x^2+11x-185=0

a = 6; b = 11; c = -185;
Δ = b2-4ac
Δ = 112-4·6·(-185)
Δ = 4561
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{-(11)-\sqrt{4561}}{2*6}=\frac{-11-\sqrt{4561}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{4561}}{2*6}=\frac{-11+\sqrt{4561}}{12}
$