Solve the equation ${x^2} + 6x - 10 = 0$

Hint: In the given question we are asked to solve the quadratic equation, that is we need to find the roots of the given quadratic equation. Since it is an equation of order two it will have two roots. The roots of a quadratic equation can be found by using the formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . Let \[a{x^2} + bx + c = 0\] be a quadratic equation, by taking the coefficient of ${x^2}$ as $a$ , and coefficient of $x$ as b and constant as $c$ .

Complete answer:
It is given that ${x^2} + 6x - 10 = 0$ we aim to solve this equation that is we have to find its roots.
We know that the number of roots of an equation is equal to its degree. Here the degree of the given equation is two thus this equation will have two roots.
The roots of a quadratic equation can be found by using the formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] where \[a\] - coefficient of the term \[{x^2}\], \[b\] - coefficient of the term \[x\], and \[c\] - constant term.
First, let us collect the required terms for the formula from the given quadratic equation to solve it.
From the given equation ${x^2} + 6x - 10 = 0$, we have \[a = 1\], \[b = 6\], and \[c = - 10\].
On substituting these terms in the formula, we get
\[x = \dfrac{{ - \left( 6 \right) \pm \sqrt {{{\left( 6 \right)}^2} - 4\left( 1 \right)\left( { - 10} \right)} }}{{2\left( 1 \right)}}\]
On simplifying this we get
\[x = \dfrac{{ - 6 \pm \sqrt {36 + 40} }}{2}\]
On further simplification we get
\[x = \dfrac{{ - 6 \pm \sqrt {76} }}{2}\] and thus we get \[x = \dfrac{{ - 6 \pm \sqrt {19 \times 4} }}{2} \Rightarrow x = \dfrac{{ - 6 \pm 2\sqrt {19} }}{2}\]
\[x = \dfrac{{ - 6 \pm 2\sqrt {19} }}{2}\]
\[ \Rightarrow x = \dfrac{{ - 6 + 2\sqrt {19} }}{2}\] and \[x = \dfrac{{ - 6 - 2\sqrt {19} }}{2}\]
On solving the above, we get
\[x = - 3 + \sqrt {19} \] and \[x = - 3 - \sqrt {19} \]
Thus, we got the roots of the given quadratic equation that is \[x = - 3 + \sqrt {19} \] and \[x = - 3 - \sqrt {19} \]

Note:
The possible values of the unknown variable in the equation is known as the roots of the equation.
The number of roots of an equation depends on its degree. The degree of an equation is the highest power of the unknown variable in that equation.
If $a = 0$ is it will form linear equation. Hence it is not possible.