How do you factor and solve ${x^2} + 6x - 7 = 0$?

Hint: In this problem we have a quadratic equation with some unknown $x$. And we are asked to factor and solve the given quadratic equation. We can factor and solve the given quadratic equation by factoring and then we can find the value of $x$. Suppose we cannot factorize the given equation then we have to use the quadratic formula to solve the given equation.

Formula used: Quadratic formula:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where $a,b,c$ are constants and $a \ne 0$, $x$ is the unknown.

Complete step-by-step solution:
Given quadratic equation is ${x^2} + 6x - 7 = 0$
To solve this given quadratic equation we are going to use a quadratic formula. A quadratic formula can be used only when the given quadratic equation is equal to zero.
General form of a quadratic equation is $a{x^2} + bx + c = 0$, now compare this with the given equation.
Then $a = 1,b = 6,c = - 7$, apply these values in the quadratic formula.
We get, $x = \dfrac{{ - 6 \pm \sqrt {{6^2} - 4.1\left( { - 7} \right)} }}{{2.1}}$, simplifying this we get,
Evaluating the exponent, we get
$ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt {36 - 4.\left( { - 7} \right)} }}{2}$,
On simplifying the numerator we get
$ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt {36 + 28} }}{2}$
Let us add the term and we get
$ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt {64} }}{2}$,
Let us taking the square root in the numerator, we get
$ \Rightarrow x = \dfrac{{ - 6 \pm 8}}{2}$
To solve for the unknown variable, separate into two equations, one with a plus symbol and the other with a minus symbol.
$ \Rightarrow x = \dfrac{{ - 6 + 8}}{2}$and
$ \Rightarrow x = \dfrac{{ - 6 - 8}}{2}$,
On solving the equations we get
$x = \dfrac{2}{2}$ and $x = \dfrac{{ - 14}}{2}$
$x = 1$ and $x = - 7$

Therefore x can take the values of 1 and -7.

Note: We can also factorize and solve the given quadratic equation by taking out the common factors.
For, the given equation is ${x^2} + 6x - 7 = 0$
We can write the middle term in the right hand side of the given equation as
${x^2} - x + 7x - 7 = 0 - - - - - (1)$
Equation (1) can be written as
$ \Rightarrow x\left( {x - 1} \right) + 7\left( {x - 1} \right) = 0$
Taking out the common terms in the above equation, we get
$ \Rightarrow \left( {x - 1} \right)\left( {x + 7} \right) = 0$
On rewriting the term,
$ \Rightarrow \left( {x - 1} \right) = 0,\left( {x + 7} \right) = 0$
$ \Rightarrow x = 1,x = - 7$