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t^2-24t-25=0
a = 1; b = -24; c = -25;
Δ = b2-4ac
Δ = -242-4·1·(-25)
Δ = 676
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{676}=26$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-26}{2*1}=\frac{-2}{2} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+26}{2*1}=\frac{50}{2} =25 $
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