b^2+3b+4=2

Solution for b^2+3b+4=2 equation:

b^2+3b+4=2

We move all terms to the left:

b^2+3b+4-(2)=0

We add all the numbers together, and all the variables

b^2+3b+2=0

a = 1; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·1·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-1}{2*1}=\frac{-4}{2}
=-2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+1}{2*1}=\frac{-2}{2}
=-1
$