which are the solutions of the quadratic equation?\n$x^{2}=-5x - 3$\n-5, 0\n$\frac{-5-sqrt{13}}{2}$…

which are the solutions of the quadratic equation?\n$x^{2}=-5x - 3$\n-5, 0\n$\frac{-5-sqrt{13}}{2}$, $\frac{-5 + sqrt{13}}{2}$\n$\frac{5-sqrt{13}}{2}$, $\frac{5+sqrt{13}}{2}$\n5, 0

which are the solutions of the quadratic equation?\n$x^{2}=-5x - 3$\n-5, 0\n$\frac{-5-sqrt{13}}{2}$, $\frac{-5 + sqrt{13}}{2}$\n$\frac{5-sqrt{13}}{2}$, $\frac{5+sqrt{13}}{2}$\n5, 0

Answer

Explanation:

Step1: Rewrite in standard form

$x^{2}+5x + 3=0$

Step2: Identify coefficients

For $ax^{2}+bx + c = 0$, here $a = 1$, $b = 5$, $c = 3$.

Step3: Use quadratic formula

$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-5\pm\sqrt{5^{2}-4\times1\times3}}{2\times1}=\frac{-5\pm\sqrt{25 - 12}}{2}=\frac{-5\pm\sqrt{13}}{2}$

Answer:

$\frac{-5 - \sqrt{13}}{2},\frac{-5+\sqrt{13}}{2}$