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3t^2+12t=9
We move all terms to the left:
3t^2+12t-(9)=0
a = 3; b = 12; c = -9;
Δ = b2-4ac
Δ = 122-4·3·(-9)
Δ = 252
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{7}}{2*3}=\frac{-12-6\sqrt{7}}{6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{7}}{2*3}=\frac{-12+6\sqrt{7}}{6} $
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