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4t^2-13t-10=0
a = 4; b = -13; c = -10;
Δ = b2-4ac
Δ = -132-4·4·(-10)
Δ = 329
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{-(-13)-\sqrt{329}}{2*4}=\frac{13-\sqrt{329}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{329}}{2*4}=\frac{13+\sqrt{329}}{8} $
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