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t^2+t-4.91=0
a = 1; b = 1; c = -4.91;
Δ = b2-4ac
Δ = 12-4·1·(-4.91)
Δ = 20.64
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{-(1)-\sqrt{20.64}}{2*1}=\frac{-1-\sqrt{20.64}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{20.64}}{2*1}=\frac{-1+\sqrt{20.64}}{2} $
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