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5t^2-19t-8=0
a = 5; b = -19; c = -8;
Δ = b2-4ac
Δ = -192-4·5·(-8)
Δ = 521
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{-(-19)-\sqrt{521}}{2*5}=\frac{19-\sqrt{521}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{521}}{2*5}=\frac{19+\sqrt{521}}{10} $
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