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8t^2+32t+24=0
a = 8; b = 32; c = +24;
Δ = b2-4ac
Δ = 322-4·8·24
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16}{2*8}=\frac{-48}{16} =-3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16}{2*8}=\frac{-16}{16} =-1 $
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