Question

Solve $x^2-[2x-8x-4]=\frac{4x-27}{3}-3$ and check your answer.

Answer 1

Given,

$x^2-(2x-8x-4)=\frac{4x-27}{3}-3$

$3x^2-3(-6x-4)=4x-27-9$

$3x^2+14x+48=0$

$a=3,b=14,c=48$

$x=\frac{-14\pm \sqrt{{14}^2-4\cdot 3\cdot 48}}{2\cdot 3}$

$=\frac{-14\pm 2\sqrt{95}i}{6}$

$x=\frac{-7\pm \sqrt{95}i}{3}$