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=-2t^2+6t+12
We move all terms to the left:
-(-2t^2+6t+12)=0
We get rid of parentheses
2t^2-6t-12=0
a = 2; b = -6; c = -12;
Δ = b2-4ac
Δ = -62-4·2·(-12)
Δ = 132
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{33}}{2*2}=\frac{6-2\sqrt{33}}{4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{33}}{2*2}=\frac{6+2\sqrt{33}}{4} $
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