b^2+16b+55=0

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Solution for b^2+16b+55=0 equation: 



b^2+16b+55=0

a = 1; b = 16; c = +55;
Δ = b2-4ac
Δ = 162-4·1·55
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-6}{2*1}=\frac{-22}{2}
=-11
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+6}{2*1}=\frac{-10}{2}
=-5
$

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