which are the solutions of $x^{2}=-11x + 4$?\n$\frac{-11-sqrt{137}}{2},\frac{-11+sqrt{137}}{2}$\n$\frac{-11-s…

which are the solutions of $x^{2}=-11x + 4$?\n$\frac{-11-sqrt{137}}{2},\frac{-11+sqrt{137}}{2}$\n$\frac{-11-sqrt{125}}{2},\frac{-11+sqrt{125}}{2}$\n$\frac{11-sqrt{137}}{2},\frac{11+sqrt{137}}{2}$\n$\frac{11-sqrt{125}}{2},\frac{11+sqrt{125}}{2}$

which are the solutions of $x^{2}=-11x + 4$?\n$\frac{-11-sqrt{137}}{2},\frac{-11+sqrt{137}}{2}$\n$\frac{-11-sqrt{125}}{2},\frac{-11+sqrt{125}}{2}$\n$\frac{11-sqrt{137}}{2},\frac{11+sqrt{137}}{2}$\n$\frac{11-sqrt{125}}{2},\frac{11+sqrt{125}}{2}$

Answer

Explanation:

Step1: Rearrange to standard quadratic form

First, rewrite the equation $x^{2}=-11x + 4$ as $x^{2}+11x - 4=0$. For a quadratic equation $ax^{2}+bx + c = 0$, here $a = 1$, $b = 11$, and $c=-4$.

Step2: Apply quadratic formula

The quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute $a = 1$, $b = 11$, and $c=-4$ into it. Calculate $b^{2}-4ac=(11)^{2}-4\times1\times(-4)=121 + 16=137$. Then $x=\frac{-11\pm\sqrt{137}}{2}$.

Answer:

A. $\frac{-11-\sqrt{137}}{2},\frac{-11 + \sqrt{137}}{2}$