3x(2+x)=33

Solution for 3x(2+x)=33 equation:

3x(2+x)=33

We move all terms to the left:

3x(2+x)-(33)=0

We add all the numbers together, and all the variables

3x(x+2)-33=0

We multiply parentheses

3x^2+6x-33=0

a = 3; b = 6; c = -33;
Δ = b2-4ac
Δ = 62-4·3·(-33)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12\sqrt{3}}{2*3}=\frac{-6-12\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+12\sqrt{3}}{2*3}=\frac{-6+12\sqrt{3}}{6}
$