4b^2+52b-169=0

Solution for 4b^2+52b-169=0 equation:

4b^2+52b-169=0

a = 4; b = 52; c = -169;
Δ = b2-4ac
Δ = 522-4·4·(-169)
Δ = 5408
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:

$\sqrt{\Delta}=\sqrt{5408}=\sqrt{2704*2}=\sqrt{2704}*\sqrt{2}=52\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-52\sqrt{2}}{2*4}=\frac{-52-52\sqrt{2}}{8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+52\sqrt{2}}{2*4}=\frac{-52+52\sqrt{2}}{8}
$