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t^2-8t-48=0
a = 1; b = -8; c = -48;
Δ = b2-4ac
Δ = -82-4·1·(-48)
Δ = 256
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{256}=16$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*1}=\frac{-8}{2} =-4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*1}=\frac{24}{2} =12 $
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