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8t^2+2t-15=0
a = 8; b = 2; c = -15;
Δ = b2-4ac
Δ = 22-4·8·(-15)
Δ = 484
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{484}=22$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-22}{2*8}=\frac{-24}{16} =-1+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22}{2*8}=\frac{20}{16} =1+1/4 $
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