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t^2+11t-13=0
a = 1; b = 11; c = -13;
Δ = b2-4ac
Δ = 112-4·1·(-13)
Δ = 173
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{173}}{2*1}=\frac{-11-\sqrt{173}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{173}}{2*1}=\frac{-11+\sqrt{173}}{2} $
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