4b^2+24b+32=0

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Solution for 4b^2+24b+32=0 equation: 



4b^2+24b+32=0

a = 4; b = 24; c = +32;
Δ = b2-4ac
Δ = 242-4·4·32
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{-(24)-8}{2*4}=\frac{-32}{8}
=-4
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8}{2*4}=\frac{-16}{8}
=-2
$

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