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t^2+5t-80=0
a = 1; b = 5; c = -80;
Δ = b2-4ac
Δ = 52-4·1·(-80)
Δ = 345
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{-(5)-\sqrt{345}}{2*1}=\frac{-5-\sqrt{345}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{345}}{2*1}=\frac{-5+\sqrt{345}}{2} $
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