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8t(t+2)=2t-5
We move all terms to the left:
8t(t+2)-(2t-5)=0
We multiply parentheses
8t^2+16t-(2t-5)=0
We get rid of parentheses
8t^2+16t-2t+5=0
We add all the numbers together, and all the variables
8t^2+14t+5=0
a = 8; b = 14; c = +5;
Δ = b2-4ac
Δ = 142-4·8·5
Δ = 36
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{36}=6$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-6}{2*8}=\frac{-20}{16} =-1+1/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+6}{2*8}=\frac{-8}{16} =-1/2 $
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