B+b2+b=12

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Solution for B+b2+b=12 equation:



+B2+B=12
We move all terms to the left:
+B2+B-(12)=0
We add all the numbers together, and all the variables
B^2+B-12=0
a = 1; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·1·(-12)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*1}=\frac{-8}{2} =-4 $
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*1}=\frac{6}{2} =3 $

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