168=(3x+3)*x

Solution for 168=(3x+3)*x equation:

168=(3x+3)*x

We move all terms to the left:

168-((3x+3)*x)=0


We calculate terms in parentheses: -((3x+3)*x), so:

(3x+3)*x

We multiply parentheses

3x^2+3x

Back to the equation:

-(3x^2+3x)


We get rid of parentheses

-3x^2-3x+168=0

a = -3; b = -3; c = +168;
Δ = b2-4ac
Δ = -32-4·(-3)·168
Δ = 2025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2025}=45$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-45}{2*-3}=\frac{-42}{-6}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+45}{2*-3}=\frac{48}{-6}
=-8
$