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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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