4b^2+12b+14=6

Solution for 4b^2+12b+14=6 equation:

4b^2+12b+14=6

We move all terms to the left:

4b^2+12b+14-(6)=0

We add all the numbers together, and all the variables

4b^2+12b+8=0

a = 4; b = 12; c = +8;
Δ = b2-4ac
Δ = 122-4·4·8
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{-(12)-4}{2*4}=\frac{-16}{8}
=-2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*4}=\frac{-8}{8}
=-1
$