How do you solve ${x^2} + 9x = - 7$ graphically and algebraically?

Hint: This equation is the quadratic equation. The general form of the quadratic equation is $a{x^2} + bx + c = 0$. Where ‘a’ is the coefficient of ${x^2}$, ‘b’ is the coefficient of x and ‘c’ is the constant term.
To solve this equation, we will apply the quadratic formula for the quadratic equation.
The formula is as below:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Here, $\sqrt {{b^2} - 4ac} $ is called the discriminant. And it is denoted by $\Delta $.
If $\Delta $ is greater than 0, then we will get two distinct and real roots.
If $\Delta $is less than 0, then we will not get real roots. In this case, we will get two complex numbers.
If $\Delta $ is equal to 0, then we will get two equal real roots.

Complete step by step solution:
Here, the quadratic equation is
$ \Rightarrow {x^2} + 9x = - 7$
Let us add 7 both sides.
$ \Rightarrow {x^2} + 9x + 7 = - 7 + 7$
That is equal to,
$ \Rightarrow {x^2} + 9x + 7 = 0$
Let us compare the above expression with $a{x^2} + bx + c = 0$.
Here, we get the value of ‘a’ is 1, the value of ‘b’ is 9, and the value of ‘c’ is 7.
Now, let us find the discriminant $\Delta $.
$ \Rightarrow \Delta = {b^2} - 4ac$
Let us substitute the values.
$ \Rightarrow \Delta = {\left( 9 \right)^2} - 4\left( 1 \right)\left( 7 \right)$
Simplify it.
$ \Rightarrow \Delta = 81 - 28$
Subtract the right-hand side.
$ \Rightarrow \Delta = 53$
Here, $\Delta $is greater than 0, then we will get two different real roots.
Now,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Put all the values.
$ \Rightarrow x = \dfrac{{ - \left( 9 \right) \pm \sqrt {53} }}{{2\left( 1 \right)}}$
That is equal to
$ \Rightarrow x = \dfrac{{ - 9 \pm \sqrt {53} }}{2}$
Hence, the two factors are $\dfrac{{ - 9 + \sqrt {53} }}{2}$ and $\dfrac{{ - 9 - \sqrt {53} }}{2}$.
Solving the equation graphically would be inefficient since the algebraic method is required to use the graphical method. However, if you are given a graph with the exact value of the x-intercepts, the x-intercepts will be the solutions to the equation.

Note:
One important thing is, we can always check our work by multiplying out factors back together, and check that we have got back the original answer.
Here is a list of methods to solve quadratic equations:
Factorization
Completing the square
Using graph
Quadratic formula