3x^2+3-3x-3=72

Solution for 3x^2+3-3x-3=72 equation:

3x^2+3-3x-3=72

We move all terms to the left:

3x^2+3-3x-3-(72)=0

We add all the numbers together, and all the variables

3x^2-3x-72=0

a = 3; b = -3; c = -72;
Δ = b2-4ac
Δ = -32-4·3·(-72)
Δ = 873
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:

$\sqrt{\Delta}=\sqrt{873}=\sqrt{9*97}=\sqrt{9}*\sqrt{97}=3\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{97}}{2*3}=\frac{3-3\sqrt{97}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{97}}{2*3}=\frac{3+3\sqrt{97}}{6}
$