Which equation could be solved using this application of the quadratic formula? -2 + 22 – 4(1)(-4)
2(1)
I

C. x^2 + 2x - 1 = 3

Step-by-step explanation:

The standard form of a quadratic equation is

ax^2 + bx + c = 0

We need to use the quadratic formula and the given expression to find the values of a, b, and c.

The quadratic formula is

Where the formula has -b, the problem has -2, so b = 2.

Now we have

ax^2 + 2x + c = 0

In the denominator, where the formula has 2a, the problem has 2(1), so a = 1.

Now we have

x^2 + 2x + c = 0

Inside the root, the quadratic formula has -4ac. the problem shows -4(1)(-4). Since we already know that a = 1, then c = -4.

Now we have

x^2 + 2x - 4 = 0

Let's look at choice A.

x^2 + 1 = 2x - 3

Subtract 2x from both sides. Add 3 to both sides.

x^2 - 2x + 4 = 0   This is not it!

Let's look at choice B.

x^2 - 2x - 1 = 3

Subtract 3 from both sides.

x^2 - 2x - 4 = 0     This is not it!

Let's look at choice C.

x^2 + 2x - 1 = 3

Subtract 3 from both sides.

x^2 + 2x - 4 = 0     This is it!

C. x^2 + 2x - 1 = 3

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.