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t^2+15t-448=0
a = 1; b = 15; c = -448;
Δ = b2-4ac
Δ = 152-4·1·(-448)
Δ = 2017
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{2017}}{2*1}=\frac{-15-\sqrt{2017}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{2017}}{2*1}=\frac{-15+\sqrt{2017}}{2} $
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