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8t^2-16t-24=0
a = 8; b = -16; c = -24;
Δ = b2-4ac
Δ = -162-4·8·(-24)
Δ = 1024
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{1024}=32$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-32}{2*8}=\frac{-16}{16} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+32}{2*8}=\frac{48}{16} =3 $
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