Answer
no real solution
Work Step by Step
Factoring the $GCF=15,$ the given equation, $
15x^2+45=0
,$ is equivalent to
\begin{align*}
15(x^2+3)&=0
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
\dfrac{\cancel{15}(x^2+3)}{\cancel{15}}&=\dfrac{0}{15}
\\\\
x^2+3&=0
\\
x^2+3-3&=0-3
\\
x^2&=-3
.\end{align*}
Taking the square root of both sides (Square Root Principle), the equation above is equivalent to
\begin{align*}
x&=\pm\sqrt{-3}
.\end{align*}
Since the radicand (i.e. $-3$) of the square root sign is negative, the solution is not a real number. Hence, the equation $15x^2+45=0$ has no real solution.
