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15t-2t^2-25=0
a = -2; b = 15; c = -25;
Δ = b2-4ac
Δ = 152-4·(-2)·(-25)
Δ = 25
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{25}=5$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5}{2*-2}=\frac{-20}{-4} =+5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5}{2*-2}=\frac{-10}{-4} =2+1/2 $
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