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22t^2+25t=0
a = 22; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·22·0
Δ = 625
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{625}=25$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*22}=\frac{-50}{44} =-1+3/22 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*22}=\frac{0}{44} =0 $
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