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8t^2+22t+15=0
a = 8; b = 22; c = +15;
Δ = b2-4ac
Δ = 222-4·8·15
Δ = 4
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{4}=2$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2}{2*8}=\frac{-24}{16} =-1+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2}{2*8}=\frac{-20}{16} =-1+1/4 $
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