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10t-10t^2=0
a = -10; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-10)·0
Δ = 100
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{100}=10$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*-10}=\frac{-20}{-20} =1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*-10}=\frac{0}{-20} =0 $
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