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t2+5t+1=0
We add all the numbers together, and all the variables
t^2+5t+1=0
a = 1; b = 5; c = +1;
Δ = b2-4ac
Δ = 52-4·1·1
Δ = 21
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{21}}{2*1}=\frac{-5-\sqrt{21}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{21}}{2*1}=\frac{-5+\sqrt{21}}{2} $
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