b^2-4=3

Solution for b^2-4=3 equation:

b^2-4=3

We move all terms to the left:

b^2-4-(3)=0

We add all the numbers together, and all the variables

b^2-7=0

a = 1; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·1·(-7)
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{-(0)-2\sqrt{7}}{2*1}=\frac{0-2\sqrt{7}}{2}
=-\frac{2\sqrt{7}}{2}
=-\sqrt{7}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{7}}{2*1}=\frac{0+2\sqrt{7}}{2}
=\frac{2\sqrt{7}}{2}
=\sqrt{7}
$