s^2-11s+28=0

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Solution for s^2-11s+28=0 equation: 



s^2-11s+28=0

a = 1; b = -11; c = +28;
Δ = b2-4ac
Δ = -112-4·1·28
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3}{2*1}=\frac{8}{2}
=4
$


$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3}{2*1}=\frac{14}{2}
=7
$

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