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-33t+5t^2-56=0
a = 5; b = -33; c = -56;
Δ = b2-4ac
Δ = -332-4·5·(-56)
Δ = 2209
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2209}=47$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-47}{2*5}=\frac{-14}{10} =-1+2/5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+47}{2*5}=\frac{80}{10} =8 $
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