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-16t^2+14t=0
a = -16; b = 14; c = 0;
Δ = b2-4ac
Δ = 142-4·(-16)·0
Δ = 196
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{196}=14$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14}{2*-16}=\frac{-28}{-32} =7/8 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14}{2*-16}=\frac{0}{-32} =0 $
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