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]
