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-12x+6=-3x^2+9
We move all terms to the left:
-12x+6-(-3x^2+9)=0
We get rid of parentheses
3x^2-12x-9+6=0
We add all the numbers together, and all the variables
3x^2-12x-3=0
a = 3; b = -12; c = -3;
Δ = b2-4ac
Δ = -122-4·3·(-3)
Δ = 180
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{5}}{2*3}=\frac{12-6\sqrt{5}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{5}}{2*3}=\frac{12+6\sqrt{5}}{6} $
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