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5t^2-11t-3=0
a = 5; b = -11; c = -3;
Δ = b2-4ac
Δ = -112-4·5·(-3)
Δ = 181
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{-(-11)-\sqrt{181}}{2*5}=\frac{11-\sqrt{181}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{181}}{2*5}=\frac{11+\sqrt{181}}{10} $
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