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3t^2+9-12t=0
a = 3; b = -12; c = +9;
Δ = b2-4ac
Δ = -122-4·3·9
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6}{2*3}=\frac{6}{6} =1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6}{2*3}=\frac{18}{6} =3 $
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