b^2-19b+68=0

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Solution for b^2-19b+68=0 equation: 



b^2-19b+68=0

a = 1; b = -19; c = +68;
Δ = b2-4ac
Δ = -192-4·1·68
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{89}}{2*1}=\frac{19-\sqrt{89}}{2}
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{89}}{2*1}=\frac{19+\sqrt{89}}{2}
$

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