Let's start by rewriting the equation in such way that all the terms are on the left side of the equality sign.
10+a^2=- 7a
⇕
10+a^2+7a=0
To make the process of factoring easier, let's rearrange the order of the terms in such way that they follow the standard form of a quadratic function.
10+a^2+7a=0
⇕
a^2+7a+10=0
Now, to factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's take a look at the constant term.
a^2+7a+ 10=0
In this case, we have 10. This is a positive number, so for the product of the constant terms in the factors to be positive, these constants must have the same sign (both positive or both negative.)
Factor Constants
Product of Constants
1 and 10
10
-1 and - 10
10
2 and 5
10
-2 and - 5
10
Next, let's consider the coefficient of the linear term.
a^2+ 7a+ 10=0
For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, 7.
Factors
Sum of Factors
1 and 10
11
-1 and - 10
- 11
2 and 5
7
-2 and - 5
- 7
We found the factors whose product is 10 and whose sum is 7.
a^2+ 7a+ 10=0
⇕
(a+5)(a+2)=0
Solving the Equation
Now, as we have already factored the equation, we can apply the Zero Product Property to solve it.
