Question

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

Answer 1

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

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...

Answer 2

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