3x(2x+10)+(2x+5)+(2x-15)=360

Solution for 3x(2x+10)+(2x+5)+(2x-15)=360 equation:

3x(2x+10)+(2x+5)+(2x-15)=360

We move all terms to the left:

3x(2x+10)+(2x+5)+(2x-15)-(360)=0

We multiply parentheses

6x^2+30x+(2x+5)+(2x-15)-360=0

We get rid of parentheses

6x^2+30x+2x+2x+5-15-360=0

We add all the numbers together, and all the variables

6x^2+34x-370=0

a = 6; b = 34; c = -370;
Δ = b2-4ac
Δ = 342-4·6·(-370)
Δ = 10036
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:

$\sqrt{\Delta}=\sqrt{10036}=\sqrt{4*2509}=\sqrt{4}*\sqrt{2509}=2\sqrt{2509}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-2\sqrt{2509}}{2*6}=\frac{-34-2\sqrt{2509}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+2\sqrt{2509}}{2*6}=\frac{-34+2\sqrt{2509}}{12}
$