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3t^2+12t-36=0
a = 3; b = 12; c = -36;
Δ = b2-4ac
Δ = 122-4·3·(-36)
Δ = 576
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{576}=24$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-24}{2*3}=\frac{-36}{6} =-6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+24}{2*3}=\frac{12}{6} =2 $
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