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5t^2+t-22=0
a = 5; b = 1; c = -22;
Δ = b2-4ac
Δ = 12-4·5·(-22)
Δ = 441
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{441}=21$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-21}{2*5}=\frac{-22}{10} =-2+1/5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+21}{2*5}=\frac{20}{10} =2 $
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