Since the given function is
[tex]f(x)=7x^4-17x^2-12[/tex]Take 7 as a common factor
[tex]f(x)=7(x^4-\frac{17}{7}x^2-\frac{12}{7})[/tex]Now let us make the trinomial completing square
Divide the middle term by 2, then square it
[tex]\begin{gathered} \frac{\frac{17}{7}}{2}x=\frac{17}{14} \\ (\frac{17}{14})^2=\frac{289}{196} \end{gathered}[/tex]So we will add and subtract this value
[tex]\begin{gathered} x^4-\frac{17}{7}x-12=x^4-\frac{17}{7}x^2+\frac{289}{196}-\frac{289}{196}-12 \\ =(x^4-\frac{17}{7}x^2+\frac{289}{196})-\frac{2641}{196} \end{gathered}[/tex]The trinomial is the square of
[tex](x^4-\frac{17}{7}x^2+\frac{289}{196})=(x^2-\frac{17}{14})^2[/tex]Then the new f(x) is
[tex]f(x)=7\lbrack(x^2-\frac{17}{14})^2-\frac{2641}{196}\rbrack[/tex]Now we can solve it by equating f(x) by 0
[tex]\begin{gathered} f(x)=0 \\ 7\lbrack(x^2-\frac{17}{14})^2-\frac{2641}{196}\rbrack=0 \end{gathered}[/tex]Divide both sides by 7 and add 2641/196 to both sides
[tex]\begin{gathered} (x^2-\frac{17}{14})^2-\frac{2641}{196}+\frac{2641}{196}=0+\frac{2641}{196} \\ (x^2-\frac{17}{14})^2=\frac{2641}{196} \end{gathered}[/tex]Take a square root to both sides
[tex](x^2-\frac{17}{14})=\pm\sqrt[]{\frac{2641}{196}}[/tex]Add 17/14 to both sides
[tex]\begin{gathered} x^2-\frac{17}{14}+\frac{17}{14}=\pm\sqrt[]{\frac{2641}{196}}+\frac{17}{14} \\ x^2=\pm4.885047188 \end{gathered}[/tex]We will cancel the -ve value because no square root of -ve values, then
Take a square root for 4.885047188 to find x
[tex]\begin{gathered} x=\pm\sqrt[]{4.885047188} \\ x=\pm2.210214286 \end{gathered}[/tex]