Given
[tex]\sqrt{2x+13}-5=x[/tex]We can find the value of x below
Explanation
Step 1: Solve for x
[tex]\begin{gathered} \sqrt{2x+13}-5=x \\ Add\text{ 5 to both sides} \\ \sqrt{2x+13}-5+5=x+5 \\ \sqrt{2x+13}=x+5 \\ Take\text{ the square of both sides} \\ (\sqrt{2x+13})^2=(x+5)^2 \\ 2x+13=x^2+10x+25 \\ Move\text{ terms to one side} \\ x^2+10x+25-2x-13=0 \\ x^2+8x+12=0 \\ We\text{ will solve the above using quadratic formula} \\ \mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:} \\ x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \mathrm{For\:}\quad a=1,\:b=8,\:c=12 \\ x_{1,\:2}=\frac{-8\pm \sqrt{8^2-4\cdot \:1\cdot \:12}}{2\cdot \:1} \\ x_{1,\:2}=\frac{-8\pm \:4}{2\cdot \:1} \\ \mathrm{Separate\:the\:solutions} \\ x_1=\frac{-8+4}{2\cdot \:1},\:x_2=\frac{-8-4}{2\cdot \:1} \\ \mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:} \\ x=-2,\:x=-6 \\ \end{gathered}[/tex]Step 2: Verify solutions
[tex]\begin{gathered} For\text{ x=-2} \\ \sqrt{2\left(-2\right)+13}-5=-2 \\ \sqrt{9}-5=-2 \\ 3-5=-2 \\ -2=-2;True \\ For\text{ x=-6} \\ \sqrt{2\left(-6\right)+13}-5=-6 \\ \sqrt{-12+13}-5=-6 \\ \sqrt{1}-5=-6 \\ -4\ne-6;False \end{gathered}[/tex]Answer: x = -2
