Hello there, I've got a quadratic equation. I'm supposed to solve it by factorising but it's already factorized.
$\displaystyle (x-2)(2x+1)=0$
Can you help me?
Better check my workings but,
$\displaystyle (3x + 1)(x + 3) = 19 - 2(x +2)$
$\displaystyle 3x^2+9x+x+3 = 19 - 2x -4$
$\displaystyle 3x^2+10x+3 = 15 - 2x$
$\displaystyle 3x^2+12x-12 = 0$
$\displaystyle 3(x^2+4x-4) = 0$
$\displaystyle x^2+4x-4 = 0$