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15t^2-16t=0
a = 15; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·15·0
Δ = 256
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{256}=16$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*15}=\frac{0}{30} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*15}=\frac{32}{30} =1+1/15 $
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