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15=3x^2-12x
We move all terms to the left:
15-(3x^2-12x)=0
We get rid of parentheses
-3x^2+12x+15=0
a = -3; b = 12; c = +15;
Δ = b2-4ac
Δ = 122-4·(-3)·15
Δ = 324
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{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-18}{2*-3}=\frac{-30}{-6} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+18}{2*-3}=\frac{6}{-6} =-1 $
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