b^2-2b-65=0

Solution for b^2-2b-65=0 equation:

b^2-2b-65=0

a = 1; b = -2; c = -65;
Δ = b2-4ac
Δ = -22-4·1·(-65)
Δ = 264
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:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{66}}{2*1}=\frac{2-2\sqrt{66}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{66}}{2*1}=\frac{2+2\sqrt{66}}{2}
$