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18=3x^2-12x
We move all terms to the left:
18-(3x^2-12x)=0
We get rid of parentheses
-3x^2+12x+18=0
a = -3; b = 12; c = +18;
Δ = b2-4ac
Δ = 122-4·(-3)·18
Δ = 360
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{10}}{2*-3}=\frac{-12-6\sqrt{10}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{10}}{2*-3}=\frac{-12+6\sqrt{10}}{-6} $
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