Respuesta:
[tex]\left(x+1\right)^2=24\left(x+4\right)\\\\x^2+2x+1=24x+96\\\\x^2+2x-95=24x\\\\x^2-22x-95=0[/tex]
Aplicamos la formula general
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]x_{1,\:2}=\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\cdot \:1\cdot \left(-95\right)}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-22\right)\pm \:12\sqrt{6}}{2\cdot \:1}\\\\x_1=\frac{-\left(-22\right)+12\sqrt{6}}{2\cdot \:1},\:x_2=\frac{-\left(-22\right)-12\sqrt{6}}{2\cdot \:1}\\\\[/tex]
[tex]x_1=\frac{-\left(-22\right)+12\sqrt{6}}{2\cdot \:1}\\\\x_1=\frac{22+12\sqrt{6}}{2\cdot \:1}\\\\x_1=\frac{2\left(11+6\sqrt{6}\right)}{2}\\\\x_1=11+6\sqrt{6}[/tex]
[tex]x_2=\frac{-\left(-22\right)-12\sqrt{6}}{2\cdot \:1}\\\\x_2=\frac{22-12\sqrt{6}}{2\cdot \:1}\\\\x_2=\frac{2\left(11-6\sqrt{6}\right)}{2}\\\\x_2=11-6\sqrt{6}[/tex]
Las soluciones de la ecuación de segundo grado son:
[tex]x=11+6\sqrt{6},\:x=11-6\sqrt{6}[/tex]
