3x(3+x)=33

Solution for 3x(3+x)=33 equation:

3x(3+x)=33

We move all terms to the left:

3x(3+x)-(33)=0

We add all the numbers together, and all the variables

3x(x+3)-33=0

We multiply parentheses

3x^2+9x-33=0

a = 3; b = 9; c = -33;
Δ = b2-4ac
Δ = 92-4·3·(-33)
Δ = 477
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:

$\sqrt{\Delta}=\sqrt{477}=\sqrt{9*53}=\sqrt{9}*\sqrt{53}=3\sqrt{53}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{53}}{2*3}=\frac{-9-3\sqrt{53}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{53}}{2*3}=\frac{-9+3\sqrt{53}}{6}
$