y=x^2+10x+3

y=x^2+10x+3 how do you solve this using the square and turning it into vertex mode

$$\begin{array}{rcl}
y&=&x^2+10x+3\\
y&=&(x+5)^2-25+3\\
y&=&(x+5)^2-22\\
\end{array}\\\\
\rm{vertex}=(-5,-22)$$

$$\begin{array}{rcl}
0 &=& (x+5)^2-22\\
(x+5)^2 &=& 22 \qquad | \qquad \sqrt\\
x_{1,2}+5 &=& \pm\sqrt{22} \\
x_{1,2} &=& -5 \pm\sqrt{22} \\
x_1 &=& -5 + \sqrt{22} = -0.3095842402\\
x_2 &=& -5 - \sqrt{22} = -9.6904157598
\end{array}$$