b^2+3=67

Solution for b^2+3=67 equation:

b^2+3=67

We move all terms to the left:

b^2+3-(67)=0

We add all the numbers together, and all the variables

b^2-64=0

a = 1; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·1·(-64)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*1}=\frac{-16}{2}
=-8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*1}=\frac{16}{2}
=8
$