How would I solve this equation algebraically?

How would I solve this equation algebraically?
$\large{ \sqrt{5x^2+11}} = x + 5$

$\begin{array}{|rcll|} \hline \sqrt{5x^2+11} &=& x + 5 \quad & | \quad \text{square both sides} \\ 5x^2+11 &=& (x + 5)^2 \\ 5x^2+11 &=& x^2+10x+25 \\ 4x^2-10x-14 &=& 0 \quad & | \quad : 2 \\ 2x^2-5x-7 &=& 0 \\ \\ x &=& \dfrac{ 5\pm\sqrt{25-4\cdot 2 \cdot (-7) } } {2\cdot 2 } \\\\ x &=& \dfrac{ 5\pm\sqrt{25+56 } } { 4 } \\\\ x &=& \dfrac{ 5\pm\sqrt{81} } { 4 } \\\\ x &=& \dfrac{ 5\pm 9 } { 4 } \\\\ x_1 &=& \dfrac{ 5 + 9 } { 4 } \\\\ x_1 &=& \dfrac{ 14 } { 4 } \\\\ \mathbf{ x_1 } &\mathbf{ =}& \mathbf{ \dfrac{ 7 } { 2 }} \\\\ x_2 &=& \dfrac{ 5 - 9 } { 4 } \\\\ x_2 &=& -\dfrac{ 4 } { 4 } \\\\ \mathbf{ x_2} &\mathbf{ =}& \mathbf{ -1} \\ \hline \end{array}$

${ \sqrt{5x^2+11}} = x + 5$... you have this equation.

Sqaure both sides to give you $5x^2 + 11 = (x+5)^2$

Solving this, we get $5x^2 + 11 = x^2 + 10x + 25$.

This is equal to $4x^2 - 10x - 14 = 0$

Factorizing this equation, we get $(2x-7)(2x+2) = 0$. (You can test that out to see if it matches the previous equation of $4x^2 - 10x - 14 = 0$.) 

Since $(2x-7)(2x+2)$ has to equal zero, at least one of the brackets need to equal zero for the equation to equal zero. Therefore, we can end up with two solutions.

$x$ is either $\boxed{\frac{7}{2}}$, or $x$ is $\boxed{-1}$.