Question

What is the solution: (9x+2)(4x^2+35x-9)=0

Answer 1

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

Answer 2

$(9x+2)(4x^2+35x-9)=0\iff9x+2=0\ or\ 4x^2+35x-9=0\\\\(\#1)\\9x+2=0\ \ \ \ \ \ |subtract\ 2\ from\ both\ sides\\9x=-2\ \ \ \ \ \ \ |divide\ both\ sides\ by\ 9\\\boxed{x=-\frac{2}{9}}\\\\(\#2)\\4x^2+35x-9=0\\4x^2+36x-x-9=0\\4x(x+9)-1(x+9)=0\\(x+9)(4x-1)=0\iff x+9=0\ or\ 4x-1=0\\x=-9\ or\ 4x=1\\\boxed{x=-9}\ or\ \boxed{x=\frac{1}{4}}\\\\Solutions:x=-\frac{2}{9}\ or\ x=-9\ or\ x=\frac{1}{4}.$