b^2-6b-49=0

Solution for b^2-6b-49=0 equation:

b^2-6b-49=0

a = 1; b = -6; c = -49;
Δ = b2-4ac
Δ = -62-4·1·(-49)
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{58}}{2*1}=\frac{6-2\sqrt{58}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{58}}{2*1}=\frac{6+2\sqrt{58}}{2}
$