b^2-18b+21=4

Solution for b^2-18b+21=4 equation:

b^2-18b+21=4

We move all terms to the left:

b^2-18b+21-(4)=0

We add all the numbers together, and all the variables

b^2-18b+17=0

a = 1; b = -18; c = +17;
Δ = b2-4ac
Δ = -182-4·1·17
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{-(-18)-16}{2*1}=\frac{2}{2}
=1
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+16}{2*1}=\frac{34}{2}
=17
$