b^2+32=5-12b

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Solution for b^2+32=5-12b equation:



b^2+32=5-12b
We move all terms to the left:
b^2+32-(5-12b)=0
We add all the numbers together, and all the variables
b^2-(-12b+5)+32=0
We get rid of parentheses
b^2+12b-5+32=0
We add all the numbers together, and all the variables
b^2+12b+27=0
a = 1; b = 12; c = +27;
Δ = b2-4ac
Δ = 122-4·1·27
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{-(12)-6}{2*1}=\frac{-18}{2} =-9 $
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6}{2*1}=\frac{-6}{2} =-3 $

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