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x(3x+3)=36
We move all terms to the left:
x(3x+3)-(36)=0
We multiply parentheses
3x^2+3x-36=0
a = 3; b = 3; c = -36;
Δ = b2-4ac
Δ = 32-4·3·(-36)
Δ = 441
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{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*3}=\frac{-24}{6} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*3}=\frac{18}{6} =3 $
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