3x(x+10)+(x+20)+40=360

Solution for 3x(x+10)+(x+20)+40=360 equation:

3x(x+10)+(x+20)+40=360

We move all terms to the left:

3x(x+10)+(x+20)+40-(360)=0

We add all the numbers together, and all the variables

3x(x+10)+(x+20)-320=0

We multiply parentheses

3x^2+30x+(x+20)-320=0

We get rid of parentheses

3x^2+30x+x+20-320=0

We add all the numbers together, and all the variables

3x^2+31x-300=0

a = 3; b = 31; c = -300;
Δ = b2-4ac
Δ = 312-4·3·(-300)
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{-(31)-\sqrt{4561}}{2*3}=\frac{-31-\sqrt{4561}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+\sqrt{4561}}{2*3}=\frac{-31+\sqrt{4561}}{6}
$