What are the solutions to the equation? b^2 − 10b=39

The equation b^2 + 10b = 39 is similar to the 'Quadratic Equation:'

y = ax^2 + bx + c

Instead of the term x, you have replaced it with the term 'b.'

So, in theory, to find x (in this case b) we make y = 0 by inversing 39 from the right hand side (RHS) to the left hand side (LHS):

b^2 + 10b - 39 = 0

Notice the similarity between this equation and the original quadratic equation:

b^2 + 10b - 39 = 0
ax^2 + bx + c = y

The term a = 1 for your formula, b = 10, and c = -39. The term x = b = ?, and y = 0.

*From now on, I'll use the original quadratic formula terms. So your formula will look like this:

x^2 + 10x - 39 = 0

We can find the solution to x through two methods: sum and product, or the quadratic formula.

METHOD ONE: SUM AND PRODUCT
To find x, we find the sum and product of a × c, and b.

a × c = 1 × -39
= -39

b = 10

Now we need two numbers that have a sum of 10 and a product of -39.

The two numbers are 13 and -3. Proof:

13 × -3
= -39 {which is the product}

13 + -3
= 13 -3 {a + and - make a -}
= 10 {which is the sum}

Now that we know these are the two numbers, we rewrite the equation like so:

(x + 13) (x - 3) = 0

Now, to find x, we know that either (x + 13) or (x - 3) equals 0, because the value x are the x-intercepts of the graph. And the x-intercepts are where y = 0. So:

x + 13 = 0

OR

x - 3 = 0

Now we inverse the values that we do know from the LHS to the RHS (see above for definitions of LHS and RHS):

x = -13

OR

x = 3

So the values of x (or the x-intercepts) are -13 and 3. So for your case b = -13 and 3.

We can prove this by replacing x in the formula:

x^2 + 10x - 39
= (-13)^2 + 10(-13) - 39
= 169 + (-130) - 39
= 169 - 130 - 39
= 0 {therefore -13 is a solution of this equation}

OR

x^2 + 10x - 39
= (3)^2 + 10(3) - 39
= 9 + 30 - 39
= 0 {therefore 3 is a solution of this equation}

METHOD TWO: QUADRATIC FORMULA
The Quadratic Formula is very easy to use once you know it's structure:

x = (-b +/- _/b^2 - 4ac) / 2a

From here, all you have to do is substitute the variables:

a = 1
b = 10
c = -39

*variables from original quadratic equation

Therefore:

x = (-10 +/- _/10^2 - 4(1)(-39)) / 2(1)
= (-10 +/- _/100 - (-156)) / 2
= (-10 +/- _/100 + 156) / 2
= (-10 +/- _/256) / 2
= (-10 +/- 16) / 2

Now, the +/- (plus or minus) symbol diverts your options into two. We can find one value of x by adding -10 and 16, and another value of x by subtracting 16 from -10:

x = (-10 + 16) / 2
= 6 / 2
= 3 {which we know is a solution of the equation}

OR

x = (-10 - 16) / 2
= (-26) / 2
= -13 {which we know is another solution of the equation}

Therefore, the solutions to b (in terms of your equation) are - 13 and 3.

I’m not gonna do that