[tex]Factoring x^2+20x+50 \\ \\ x^2 + 20x + 50 = 0 \\ \\ we \ the \ find \ the \ vertex \ of \ y = x^2+20x+50 \\ \\ Since \ a \ square \ root \ has \ two \ values, \ one \\ positive \ and \ the \ other \ negative. \\ \\ x^2 + 20x + 50 = 0 \\has \ (2) \ solutions \\ x = -10 (+) \sqrt 50... \\ or... \\ x = -10 (-) \sqrt 50 \\ \\ [/tex]
Sense the solution would then equal 100, we would then start switching.
10= 100
5= 50.
[tex] \left[\begin{array}{ccc}100\div5=20\end{array}\right] [/tex]
In this case, 20 would equal 2.
Your answer would then be the last option.
