Given:
[tex]x^2-5x-24=0[/tex]First, let us keep the x terms on the left side and move the constant to the right side of the equation through adding 24 on both sides.
[tex]x^2-5x-24+24=0+24[/tex][tex]x^2-5x=24[/tex]Next, we will take the half of the x term, 5x, and square it.
[tex](\frac{5}{2})^2=\frac{25}{4}[/tex]And then, we will add this to both sides of the equation.
[tex]x^2-5x+\frac{25}{4}=24+\frac{25}{4}[/tex]Then, we will rewrite the left side of the equation as a perfect square
[tex](x-\frac{5}{2})^2=\frac{121}{4}[/tex]Take the square root of both sides
[tex]\sqrt[]{(x-\frac{5}{2})^2}=\sqrt[]{\frac{121}{4}}[/tex][tex]x-\frac{5}{2}=\pm\frac{11}{2}[/tex]Solve for x (1)
[tex]x=\frac{5}{2}+\frac{11}{2}[/tex][tex]x=\frac{16}{2}[/tex][tex]x=8[/tex]Solve for x (2)
[tex]x=\frac{5}{2}-\frac{11}{2}[/tex][tex]x=\frac{-6}{2}[/tex][tex]x=-3[/tex]Now, we know that the values of x are -3 and 8, therefore, the answer would be C. x=-3 and x=8.