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2t^2+9t-200=0
a = 2; b = 9; c = -200;
Δ = b2-4ac
Δ = 92-4·2·(-200)
Δ = 1681
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{1681}=41$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-41}{2*2}=\frac{-50}{4} =-12+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+41}{2*2}=\frac{32}{4} =8 $
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