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3t^2+18t+15=0
a = 3; b = 18; c = +15;
Δ = b2-4ac
Δ = 182-4·3·15
Δ = 144
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{144}=12$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-12}{2*3}=\frac{-30}{6} =-5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+12}{2*3}=\frac{-6}{6} =-1 $
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