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-3q^2+12=0
a = -3; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-3)·12
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-3}=\frac{-12}{-6} =+2 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-3}=\frac{12}{-6} =-2 $
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