Solve a rational equation Precalculus

Solve the following equations.

\dfrac{x^{2}+2x-3}{x-4}=0

x=1 and x=-3

x=-1 and x=3

x=3

x=\dfrac{3-\sqrt{33}}{2} and x=\dfrac{3+\sqrt{33}}{2}

\dfrac{4x^2+5x-10}{3x+5}+\dfrac{2x^2-4x-5}{3x+5}=0

x=-\dfrac{3}{2} and x=\dfrac{5}{3}

x=\dfrac{1}{2} and x=-\dfrac{3}{2}

x=\dfrac{3}{4} and x=-\dfrac{7}{4}

x=\dfrac{3}{2}

\dfrac{2x^2+8x-10}{x-1}=3

x=\dfrac{7}{2} and x=-1

x=-\dfrac{5}{2} and x=2

x=-\dfrac{7}{2}

x=\dfrac{5}{2} and x=-1

\dfrac{2x^2-5x-25}{2x+5}=0

x=\dfrac{1}{2} and x=-5

x=-5

x=\dfrac{5}{2} and x=-5

x=5

\dfrac{7x^2+2x-1}{4x-5}+\dfrac{x^2-16x+6}{4x-5}=0

x=\dfrac{1}{2}

x=-\dfrac{1}{2} and x=-\dfrac{5}{4}

x=-\dfrac{1}{2} and x=5

x=-\dfrac{1}{2} and x=\dfrac{3}{2}

\dfrac{4x^2+4x-24}{x+3}=12

x=5

x=-5 and x=3

x=-5 and x=\dfrac{3}{8}

x=-5 and x=\dfrac{3}{4}

\dfrac{6x^2+11x-7}{3x+7}=0

x=-\dfrac{1}{3} and x=\dfrac{5}{6}

x=-\dfrac{1}{2} and x=\dfrac{5}{3}

x=-\dfrac{1}{2} and x=\dfrac{7}{3}

x=\dfrac{1}{2}