Enter the correct answer in the box.

Replace the values of [tex][tex]$m$[/tex][/tex] and [tex][tex]$n$[/tex][/tex] to show the solutions of this equation.

[tex]\[ x^2 + 6x - 5 = 0 \][/tex]

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
[tex][tex]$\frac{\square}{\square} \quad$[/tex][/tex] & [tex][tex]$\pi$[/tex][/tex] & [tex][tex]$a$[/tex][/tex] & [tex][tex]$\beta$[/tex][/tex] & [tex][tex]$\varepsilon$[/tex][/tex] & 9 & [tex][tex]$\sin \cos$[/tex][/tex] & [tex][tex]$\tan$[/tex][/tex] & [tex][tex]$\sqrt{0}$[/tex][/tex] & [tex][tex]$\sqrt[5]{\square}$[/tex][/tex] & 4 & [tex][tex]$\ \textless \ $[/tex][/tex] & [tex][tex]$\ \textgreater \ $[/tex][/tex] & [tex][tex]$\leq$[/tex][/tex] & [tex][tex]$\geq$[/tex][/tex] & [tex][tex]$\lambda$[/tex][/tex] \\
\hline
[tex][tex]$\mu$[/tex][/tex] & [tex][tex]$\rho$[/tex][/tex] & [tex][tex]$\varphi$[/tex][/tex] & [tex][tex]$\csc \sec$[/tex][/tex] & [tex][tex]$\cot$[/tex][/tex] & & & & & & & & & & & \\
\hline
\end{tabular}

[tex]\[ x^2 + 6x - 5 = 0 \][/tex]



Answer :

To solve the quadratic equation [tex]\( x^2 + 6x - 5 = 0 \)[/tex], we need to find the values of [tex]\( x \)[/tex] that satisfy the equation. These values are known as the roots of the equation.

The solutions of the equation [tex]\( x^2 + 6x - 5 = 0 \)[/tex] are given using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are the coefficients of the equation [tex]\( ax^2 + bx + c = 0 \)[/tex].

Given:
- [tex]\( a = 1 \)[/tex] (coefficient of [tex]\( x^2 \)[/tex])
- [tex]\( b = 6 \)[/tex] (coefficient of [tex]\( x \)[/tex])
- [tex]\( c = -5 \)[/tex] (constant term)

The quadratic formula yields:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

Substituting [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the formula:
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{36 + 20}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{56}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm 2\sqrt{14}}{2} \][/tex]
[tex]\[ x = -3 \pm \sqrt{14} \][/tex]

Thus, the solutions are:
[tex]\[ x = -3 + \sqrt{14} \][/tex]
[tex]\[ x = -3 - \sqrt{14} \][/tex]

Therefore, the roots of the equation [tex]\( x^2 + 6x - 5 = 0 \)[/tex] are:
[tex]\[ x = -3 + \sqrt{14} \quad \text{and} \quad x = -3 - \sqrt{14} \][/tex]