Answer
$x_1=2$
$x_2=-0.75$
Work Step by Step
Let's solve $\dfrac{4}{x+2}+\dfrac{3}{2x-1}=2$. Let's find common denominator.
$$\dfrac{4(2x-1)+3(x+2)}{(x+2)(2x-1)}=2$$
$$4(2x-1)+3(x+2)=2(x+2)(2x-1)$$
$$8x-4+3x+6=2(2x^2-x+4x-2)$$
$$8x-4+3x+6=4x^2+6x-4$$
$$11x+2=4x^2+6x-4$$
$$4x^2-5x-6=0$$
Let's find the discriminant. The quadratic equation in its standard form is
$$ax^2+bx+c=0.$$
$D=b^2−4ac.$ In our case $D=25-4\cdot(-6)\cdot4=121>0.$ So we have two solutions.
$$x_1=\dfrac{−b+\sqrt D}{2a}=\dfrac{5+11}{8}=2$$
$$x_2=\dfrac{−b−\sqrt D}{2a}=\dfrac{5-11}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}=-0.75$$
As none of the solutions is a zero of the denominator of the given equation, it means both solutions are correct.
