Answer
$x_1=2$
$x_2=-3$
$x_3=\dfrac{1}{2}$
Work Step by Step
We are given the equation:
$(x-2)(2x^2+5x-3)=0$
Factor the polynomial:
$(x-2)(2x^2+6x-x-3)=0$
$(x-2)[2x(x+3)-(x+3)]=0$
$(x-2)(x+3)(2x-1)=0$
$x-2=0\Rightarrow x_1=2$
$x+3=0\Rightarrow x_2=-3$
$2x-1=0\Rightarrow x_3=\dfrac{1}{2}$
