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t^2+13t+17=0
a = 1; b = 13; c = +17;
Δ = b2-4ac
Δ = 132-4·1·17
Δ = 101
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{101}}{2*1}=\frac{-13-\sqrt{101}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{101}}{2*1}=\frac{-13+\sqrt{101}}{2} $
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