which are the solutions of the quadratic equation?\n$x^{2}=7x + 4$\n$\\frac{-7-sqrt{65}}{2},\\frac{-7+sqrt{65…

which are the solutions of the quadratic equation?\n$x^{2}=7x + 4$\n$\\frac{-7-sqrt{65}}{2},\\frac{-7+sqrt{65}}{2}$\n$-7,0$\n$\\frac{7-sqrt{65}}{2},\\frac{7+sqrt{65}}{2}$\n$7,0$

which are the solutions of the quadratic equation?\n$x^{2}=7x + 4$\n$\\frac{-7-sqrt{65}}{2},\\frac{-7+sqrt{65}}{2}$\n$-7,0$\n$\\frac{7-sqrt{65}}{2},\\frac{7+sqrt{65}}{2}$\n$7,0$

Answer

Explanation:

Step1: Rewrite in standard form

First, rewrite $x^{2}=7x + 4$ as $x^{2}-7x - 4=0$. For a quadratic equation $ax^{2}+bx + c = 0$, here $a = 1$, $b=-7$, $c=-4$.

Step2: Apply quadratic - formula

The quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute $a = 1$, $b=-7$, $c=-4$ into the formula. First, calculate the discriminant $\Delta=b^{2}-4ac=(-7)^{2}-4\times1\times(-4)=49 + 16=65$. Then $x=\frac{-(-7)\pm\sqrt{65}}{2\times1}=\frac{7\pm\sqrt{65}}{2}$.

Answer:

$\frac{7-\sqrt{65}}{2},\frac{7 + \sqrt{65}}{2}$