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-16t^2+15t+12=0
a = -16; b = 15; c = +12;
Δ = b2-4ac
Δ = 152-4·(-16)·12
Δ = 993
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{993}}{2*-16}=\frac{-15-\sqrt{993}}{-32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{993}}{2*-16}=\frac{-15+\sqrt{993}}{-32} $
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