Answer
The solution is $x=-\dfrac{5}{2}+\dfrac{\sqrt{33}}{2}$
Work Step by Step
$\log_{2}x+\log_{2}(x+5)=1$
Combine the sum on the left side of the equation as a product:
$\log_{2}x(x+5)=1$
Evaluate the product inside the $\log$ on the left side:
$\log_{2}(x^{2}+5x)=1$
Rewrite the equation in exponential form:
$2^{1}=x^{2}+5x$
Rearrange:
$x^{2}+5x-2=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
For this particular equation, $a=1$, $b=5$ and $c=-2$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-5\pm\sqrt{5^{2}-4(1)(-2)}}{2(1)}=\dfrac{-5\pm\sqrt{25+8}}{2}=...$
$...=\dfrac{-5\pm\sqrt{33}}{2}=-\dfrac{5}{2}\pm\dfrac{\sqrt{33}}{2}$
Verify the solutions by plugging them into the original equation:
$x=-\dfrac{5}{2}-\dfrac{\sqrt{33}}{2}$
$\log_{2}\Big(-\dfrac{5}{2}-\dfrac{\sqrt{33}}{2}\Big)+\log_{2}\Big(-\dfrac{5}{2}-\dfrac{\sqrt{33}}{2}+5\Big)\ne1$ False
$x=-\dfrac{5}{2}+\dfrac{\sqrt{33}}{2}$
$\log_{2}\Big(-\dfrac{5}{2}+\dfrac{\sqrt{33}}{2}\Big)+\log_{2}\Big(-\dfrac{5}{2}+\dfrac{\sqrt{33}}{2}+5\Big)=1$ True
The solution is $x=-\dfrac{5}{2}+\dfrac{\sqrt{33}}{2}$
