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3t^2+15t-18=0
a = 3; b = 15; c = -18;
Δ = b2-4ac
Δ = 152-4·3·(-18)
Δ = 441
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{441}=21$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-21}{2*3}=\frac{-36}{6} =-6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+21}{2*3}=\frac{6}{6} =1 $
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