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3x^2+6x-12=-6x+6
We move all terms to the left:
3x^2+6x-12-(-6x+6)=0
We get rid of parentheses
3x^2+6x+6x-6-12=0
We add all the numbers together, and all the variables
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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