11b^2-23=0

Solution for 11b^2-23=0 equation:

11b^2-23=0

a = 11; b = 0; c = -23;
Δ = b2-4ac
Δ = 02-4·11·(-23)
Δ = 1012
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:

$\sqrt{\Delta}=\sqrt{1012}=\sqrt{4*253}=\sqrt{4}*\sqrt{253}=2\sqrt{253}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{253}}{2*11}=\frac{0-2\sqrt{253}}{22}
=-\frac{2\sqrt{253}}{22}
=-\frac{\sqrt{253}}{11}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{253}}{2*11}=\frac{0+2\sqrt{253}}{22}
=\frac{2\sqrt{253}}{22}
=\frac{\sqrt{253}}{11}
$