s^2-8s+9=0

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Solution for s^2-8s+9=0 equation:



s^2-8s+9=0
a = 1; b = -8; c = +9;
Δ = b2-4ac
Δ = -82-4·1·9
Δ = 28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{7}}{2*1}=\frac{8-2\sqrt{7}}{2} $
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{7}}{2*1}=\frac{8+2\sqrt{7}}{2} $

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