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t^2-13t-48=0
a = 1; b = -13; c = -48;
Δ = b2-4ac
Δ = -132-4·1·(-48)
Δ = 361
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{361}=19$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-19}{2*1}=\frac{-6}{2} =-3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+19}{2*1}=\frac{32}{2} =16 $
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