y=x^2+10x+3

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

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

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