b^2=289

Solution for b^2=289 equation:

b^2=289

We move all terms to the left:

b^2-(289)=0

a = 1; b = 0; c = -289;
Δ = b2-4ac
Δ = 02-4·1·(-289)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-34}{2*1}=\frac{-34}{2}
=-17
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+34}{2*1}=\frac{34}{2}
=17
$