The equation [tex]f(x)=g(x)[/tex] is[tex]|x-4|-11=-\sqrt{5x}[/tex]Some observations:[tex]\sqrt{5x}[/tex] is defined only as long as [tex]5x\ge0[/tex], or [tex]x\ge0[/tex]wherever [tex]\sqrt{5x}[/tex] is defined, its value must be non-negative, so that [tex]-\sqrt{5x}[/tex] is never positiveby the definition of absolute value, we have [tex]|x-4|=x-4[/tex] if [tex]x\ge4[/tex], and [tex]|x-4|=-(x-4)=4-x[/tex] if [tex]x<4[/tex]. Then[tex]|x-4|-11=\begin{cases}x-15&\text{for }x\ge4\\-x-7&\text{for }x<4\end{cases}[/tex]If [tex]x<4[/tex], the equation becomes[tex]-x-7=-\sqrt{5x}\implies x+7=\sqrt{5x}[/tex]Taking the square of both sides gives[tex](x+7)^2=\left(\sqrt{5x}\right)^2\implies x^2+14x+49=5x\implies x^2+9x+49=0[/tex]but since the discriminant is [tex]9^2-4\cdot1\cdot49<0[/tex], there are no real solutions.If [tex]x\ge4[/tex], then[tex]x-15=-\sqrt{5x}[/tex]Taking squares gives[tex](x-15)^2=\left(-\sqrt{5x}\right)^2\implies x^2-30x+225=5x[/tex]and solving by the quadratic formula gives two potential solutions,[tex]x=\dfrac{35\pm5\sqrt{13}}2[/tex]which have approximate values of 8.49 and 26.51.We know for any value of [tex]x[/tex] that [tex]g(x)\le0[/tex]. We have [tex]f(8.49)\approx-6.51[/tex] and [tex]f(26.51)\approx11.51[/tex], so only the first solution 8.49 is valid.