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5t^2+22t+17=0
a = 5; b = 22; c = +17;
Δ = b2-4ac
Δ = 222-4·5·17
Δ = 144
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{144}=12$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-12}{2*5}=\frac{-34}{10} =-3+2/5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+12}{2*5}=\frac{-10}{10} =-1 $
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