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t^2+25t+24=0
a = 1; b = 25; c = +24;
Δ = b2-4ac
Δ = 252-4·1·24
Δ = 529
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{529}=23$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*1}=\frac{-48}{2} =-24 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*1}=\frac{-2}{2} =-1 $
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