12b+b^2+32=0

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Solution for 12b+b^2+32=0 equation: 



12b+b^2+32=0

a = 1; b = 12; c = +32;
Δ = b2-4ac
Δ = 122-4·1·32
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4}{2*1}=\frac{-16}{2}
=-8
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*1}=\frac{-8}{2}
=-4
$

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