b^2-18b+65=0

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Solution for b^2-18b+65=0 equation: 



b^2-18b+65=0

a = 1; b = -18; c = +65;
Δ = b2-4ac
Δ = -182-4·1·65
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-8}{2*1}=\frac{10}{2}
=5
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+8}{2*1}=\frac{26}{2}
=13
$

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