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3x^2-4=6x+12
We move all terms to the left:
3x^2-4-(6x+12)=0
We get rid of parentheses
3x^2-6x-12-4=0
We add all the numbers together, and all the variables
3x^2-6x-16=0
a = 3; b = -6; c = -16;
Δ = b2-4ac
Δ = -62-4·3·(-16)
Δ = 228
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{57}}{2*3}=\frac{6-2\sqrt{57}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{57}}{2*3}=\frac{6+2\sqrt{57}}{6} $
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