14b+b^2+40=0

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Solution for 14b+b^2+40=0 equation: 



14b+b^2+40=0

a = 1; b = 14; c = +40;
Δ = b2-4ac
Δ = 142-4·1·40
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{-(14)-6}{2*1}=\frac{-20}{2}
=-10
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+6}{2*1}=\frac{-8}{2}
=-4
$

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