(9+3x)(10+3x)=240

Solution for (9+3x)(10+3x)=240 equation:

(9+3x)(10+3x)=240

We move all terms to the left:

(9+3x)(10+3x)-(240)=0

We add all the numbers together, and all the variables

(3x+9)(3x+10)-240=0

We multiply parentheses ..

(+9x^2+30x+27x+90)-240=0

We get rid of parentheses

9x^2+30x+27x+90-240=0

We add all the numbers together, and all the variables

9x^2+57x-150=0

a = 9; b = 57; c = -150;
Δ = b2-4ac
Δ = 572-4·9·(-150)
Δ = 8649
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8649}=93$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(57)-93}{2*9}=\frac{-150}{18}
=-8+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(57)+93}{2*9}=\frac{36}{18}
=2
$