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3x+3x^2=18
We move all terms to the left:
3x+3x^2-(18)=0
a = 3; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·3·(-18)
Δ = 225
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{225}=15$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-15}{2*3}=\frac{-18}{6} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+15}{2*3}=\frac{12}{6} =2 $
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