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-5t^2+15t+200=0
a = -5; b = 15; c = +200;
Δ = b2-4ac
Δ = 152-4·(-5)·200
Δ = 4225
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{4225}=65$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-65}{2*-5}=\frac{-80}{-10} =+8 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+65}{2*-5}=\frac{50}{-10} =-5 $
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