b^2+8b+20=8

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Solution for b^2+8b+20=8 equation:



b^2+8b+20=8
We move all terms to the left:
b^2+8b+20-(8)=0
We add all the numbers together, and all the variables
b^2+8b+12=0
a = 1; b = 8; c = +12;
Δ = b2-4ac
Δ = 82-4·1·12
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{-(8)-4}{2*1}=\frac{-12}{2} =-6 $
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*1}=\frac{-4}{2} =-2 $

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