b^2+34b-289=2

Solution for b^2+34b-289=2 equation:

b^2+34b-289=2

We move all terms to the left:

b^2+34b-289-(2)=0

We add all the numbers together, and all the variables

b^2+34b-291=0

a = 1; b = 34; c = -291;
Δ = b2-4ac
Δ = 342-4·1·(-291)
Δ = 2320
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:

$\sqrt{\Delta}=\sqrt{2320}=\sqrt{16*145}=\sqrt{16}*\sqrt{145}=4\sqrt{145}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-4\sqrt{145}}{2*1}=\frac{-34-4\sqrt{145}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+4\sqrt{145}}{2*1}=\frac{-34+4\sqrt{145}}{2}
$