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+287=0

a = 1; b = 34; c = +287;
Δ = b2-4ac
Δ = 342-4·1·287
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-2\sqrt{2}}{2*1}=\frac{-34-2\sqrt{2}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+2\sqrt{2}}{2*1}=\frac{-34+2\sqrt{2}}{2}
$