20-b2=3

Solution for 20-b2=3 equation:

20-b2=3

We move all terms to the left:

20-b2-(3)=0

We add all the numbers together, and all the variables

-1b^2+17=0

a = -1; b = 0; c = +17;
Δ = b2-4ac
Δ = 02-4·(-1)·17
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{17}}{2*-1}=\frac{0-2\sqrt{17}}{-2}
=-\frac{2\sqrt{17}}{-2}
=-\frac{\sqrt{17}}{-1}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{17}}{2*-1}=\frac{0+2\sqrt{17}}{-2}
=\frac{2\sqrt{17}}{-2}
=\frac{\sqrt{17}}{-1}
$