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3x^2-6x-15=12x+12
We move all terms to the left:
3x^2-6x-15-(12x+12)=0
We get rid of parentheses
3x^2-6x-12x-12-15=0
We add all the numbers together, and all the variables
3x^2-18x-27=0
a = 3; b = -18; c = -27;
Δ = b2-4ac
Δ = -182-4·3·(-27)
Δ = 648
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18\sqrt{2}}{2*3}=\frac{18-18\sqrt{2}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18\sqrt{2}}{2*3}=\frac{18+18\sqrt{2}}{6} $
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