If you subtract 17 from both sides, the equation becomes
[tex] x^2+2x-16 = 0 [/tex]
and it's a quadratic equation in standard form, i.e. [tex] ax^2+bx+c=0 [/tex]
If you name the coefficients like this, the equation has solutions
[tex] x_{1,2} = \cfrac{-b\pm\sqrt{b^2-4ac}}{2a} [/tex]
So, if you plug your values for a,b and c you have
[tex] x_{1,2} = \cfrac{-2\pm\sqrt{4-4\cdot 1 \cdot (-16)}}{2\cdot 1} = \cfrac{-2\pm\sqrt{68}}{2}[/tex]
Since 68 = 4 x 17, we have
[tex] \sqrt{68} = \sqrt{4\times 17} = \sqrt{4}\sqrt{17} = 2\sqrt{17} [/tex]
So, the solutions can be written as
[tex] \cfrac{-2\pm2\sqrt{17}}{2} = -1\pm \sqrt{17}[/tex]
