(9x+2)(4x^2+35x-9)=0

\( (9x+2)*(4x^2+35x-9)=0 \)

we can make this.

\( 9x+2=0 \)

and

\( 4x^2+35x-9=0 \)

then.

\( 9x+2=0 \)

\( \boxed{x=-\frac{2}{9}} \)


\( 4x^2+35x-9=0 \)

\( x=\frac{-b\pm\sqrt{b^2-4*a*c}}{2*a} \)

\( x=\frac{-35\pm\sqrt{35^2-4*4*(-9)}}{2*4} \)

\( x=\frac{-35\pm\sqrt{1369}}{8} \)

\( x=\frac{-35\pm37}{8} \)

\( x_1=\frac{-35+37}{8}=\frac{1}{4} \)

\( x_2=\frac{-35-37}{8}=-9 \)

\( \boxed{\boxed{S=\{\frac{1}{4},\frac{2}{9},9\}}} \)

I’m sure about my answer, don’t matter the order.

\( (9x+2)(4x^2+35x-9)=0\\\\ 9x+2=0\\ 9x=-2\\ x=-\frac29\\ \\ 4x^2+35x-9=0\\ \Delta=1225+144=1369\ \ \sqrt{\Delta}=37\\ x_1=\frac{-35-37}{2 \cdot 4}=-9\\ \\ x_2=\frac{-35+37}{2\cdot 4}=\frac14\\ \\ x\in\{-9,\frac29,\frac14\} \)


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