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1.5t^2-26.8t-16100=0
a = 1.5; b = -26.8; c = -16100;
Δ = b2-4ac
Δ = -26.82-4·1.5·(-16100)
Δ = 97318.24
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26.8)-\sqrt{97318.24}}{2*1.5}=\frac{26.8-\sqrt{97318.24}}{3} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26.8)+\sqrt{97318.24}}{2*1.5}=\frac{26.8+\sqrt{97318.24}}{3} $
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