Answer:
[tex]x=8,\:x=-3[/tex]
Step-by-step explanation:
[tex]x^2-5x-24=0\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\mathrm{For\:}\quad a=1,\:b=-5,\:c=-24:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:1\left(-24\right)}}{2\cdot \:1}\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\x=8,\:x=-3[/tex][tex]=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:1\cdot \:24}}{2\cdot \:1}\\5+\sqrt{\left(-5\right)^2+4\cdot \:1\cdot \:24}\\5+\sqrt{121}\\=\frac{5+\sqrt{121}}{2}\\=\frac{5+11}{2}\\=\frac{16}{2}\\=8\\\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:1\cdot \left(-24\right)}}{2\cdot \:1}\\=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:1\cdot \:24}}{2\cdot \:1}\\5-\sqrt{\left(-5\right)^2+4\cdot \:1\cdot \:24}\\\left(-5\right)^2\\=25\\4\cdot \:1\cdot \:24\\=96\\=5+\sqrt{25+96}\\[/tex]
[tex]\frac{5-\sqrt{121}}{2\cdot \:1}\\\frac{5-11}{2}\\=-\frac{6}{2}\\=-3\\[/tex]
