B^2-12=14-4b^2

Solution for B^2-12=14-4b^2 equation:

^2-12=14-4B^2

We move all terms to the left:

^2-12-(14-4B^2)=0

We add all the numbers together, and all the variables

-(14-4B^2)=0

We get rid of parentheses

4B^2-14=0

a = 4; b = 0; c = -14;
Δ = b2-4ac
Δ = 02-4·4·(-14)
Δ = 224
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{14}}{2*4}=\frac{0-4\sqrt{14}}{8}
=-\frac{4\sqrt{14}}{8}
=-\frac{\sqrt{14}}{2}
$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{14}}{2*4}=\frac{0+4\sqrt{14}}{8}
=\frac{4\sqrt{14}}{8}
=\frac{\sqrt{14}}{2}
$