b^2+4b=3

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

b^2+4b=3

We move all terms to the left:

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

a = 1; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·1·(-3)
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{7}}{2*1}=\frac{-4-2\sqrt{7}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{7}}{2*1}=\frac{-4+2\sqrt{7}}{2}
$