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4t^2-29t-24=0
a = 4; b = -29; c = -24;
Δ = b2-4ac
Δ = -292-4·4·(-24)
Δ = 1225
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{1225}=35$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-35}{2*4}=\frac{-6}{8} =-3/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+35}{2*4}=\frac{64}{8} =8 $
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