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3t=2+8t-t^2
We move all terms to the left:
3t-(2+8t-t^2)=0
We get rid of parentheses
t^2-8t+3t-2=0
We add all the numbers together, and all the variables
t^2-5t-2=0
a = 1; b = -5; c = -2;
Δ = b2-4ac
Δ = -52-4·1·(-2)
Δ = 33
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{33}}{2*1}=\frac{5-\sqrt{33}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{33}}{2*1}=\frac{5+\sqrt{33}}{2} $
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