Given:
[tex]f\mleft(x\mright)=x^2+2x-15[/tex]
To find: The value of x when
[tex]f(x)=0[/tex]
Explanation:
Since,
[tex]f(x)=0[/tex]
We can write it as,
[tex]\begin{gathered} x^2+2x-15=0 \\ x^2+5x-3x-15=0 \\ x(x^{}+5)-3(x-5)=0 \\ (x+5)(x-3)=0 \\ x=-5,3 \end{gathered}[/tex]
Hence, the solution is x = -5, and 3.
Final answer: The solution is,
[tex]\mleft\lbrace-5,3\mright\rbrace[/tex]
