I'm having problem solving these two equations I will include a picture

Given:
1) The equation is,
[tex]4x^3-5x^2-196x+245=0[/tex]To solve this equation,
using synthetic division,
Now solving further,
[tex]\begin{gathered} 4x^3-5x^2-196x+245=(x-7)(4x^2+23x-35) \\ take,\text{ }4x^2+23x-35=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x_{}=\frac{-23\pm\sqrt{23^2-4\cdot\:4\left(-35\right)}}{2\cdot\:4} \\ x=\frac{-23\pm\:33}{2\cdot\:4} \\ x_{}=\frac{-23+33}{2\cdot\:4},\: x_{}=\frac{-23-33}{2\cdot\:4} \\ x=\frac{5}{4},\: x=-7 \end{gathered}[/tex]Hence, the solution of given equation is,
[tex]\begin{gathered} 4x^3-5x^2-196x+245=(x-7)(x-\frac{5}{4})(x+7) \\ \Rightarrow(x-7)(x-\frac{5}{4})(x+7)=0 \\ \Rightarrow x=\text{ 7,-7,}\frac{5}{4} \end{gathered}[/tex]2) the equation is,
[tex]9x^3+2x^2+9x+2=0[/tex]Now, factor the equation,
[tex]\begin{gathered} 9x^3+2x^2+9x+2=0 \\ x^2(9x+2)+(9x+2)=0 \\ (9x+2)(x^2+1)=0 \\ \Rightarrow9x+2=0,x^2+1=0 \\ \Rightarrow x=\frac{-2}{9}, \\ \text{and x}^2=-1 \\ x=\pm\sqrt[]{-1} \\ x=\pm i \end{gathered}[/tex]Hence, the solution of above equation is x= -2/9 , i , -i.