t^2+9t+1=0

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Solution for t^2+9t+1=0 equation: 



t^2+9t+1=0

a = 1; b = 9; c = +1;
Δ = b2-4ac
Δ = 92-4·1·1
Δ = 77
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{-(9)-\sqrt{77}}{2*1}=\frac{-9-\sqrt{77}}{2}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{77}}{2*1}=\frac{-9+\sqrt{77}}{2}
$

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