b^2=9/491

Solution for b^2=9/491 equation:

b^2=9/491

We move all terms to the left:

b^2-(9/491)=0

We add all the numbers together, and all the variables

b^2-(+9/491)=0

We get rid of parentheses

b^2-9/491=0

We multiply all the terms by the denominator


b^2*491-9=0

Wy multiply elements

491b^2-9=0

a = 491; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·491·(-9)
Δ = 17676
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:

$\sqrt{\Delta}=\sqrt{17676}=\sqrt{36*491}=\sqrt{36}*\sqrt{491}=6\sqrt{491}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{491}}{2*491}=\frac{0-6\sqrt{491}}{982}
=-\frac{6\sqrt{491}}{982}
=-\frac{3\sqrt{491}}{491}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{491}}{2*491}=\frac{0+6\sqrt{491}}{982}
=\frac{6\sqrt{491}}{982}
=\frac{3\sqrt{491}}{491}
$