Solve for x ( log base 5 of x)^2-7 log base 5 of x+12=0

Consider the form ${\displaystyle x^2+bx+c}$. Find a pair of integers whose product is ${\displaystyle c}$ and whose sum is ${\displaystyle b}$. In this case, whose product is ${\displaystyle 12}$ and whose sum is ${\displaystyle -7}$.

Write the factored form using these integers.

Add ${\displaystyle 4}$ to both sides of the equation.

Rewrite ${\displaystyle {\log }_5\left(x\right)=4}$ in exponential form using the definition of a logarithm. If ${\displaystyle x}$ and ${\displaystyle b}$ are positive real numbers and ${\displaystyle b\ne 1}$, then ${\displaystyle {\log }_b\left(x\right)=y}$ is equivalent to ${\displaystyle b^y=x}$.

Solve for ${\displaystyle x}$.

Add ${\displaystyle 3}$ to both sides of the equation.

Rewrite ${\displaystyle {\log }_5\left(x\right)=3}$ in exponential form using the definition of a logarithm. If ${\displaystyle x}$ and ${\displaystyle b}$ are positive real numbers and ${\displaystyle b\ne 1}$, then ${\displaystyle {\log }_b\left(x\right)=y}$ is equivalent to ${\displaystyle b^y=x}$.

Solve for ${\displaystyle x}$.