## anonymous 5 years ago I am confused could some explain this to me: explain how to solve a quadratic equation using the quadratic formula and demonstrate the process

1. anonymous

-b + or - $-b \pm \sqrt{b^2-4ac}$

2. anonymous

not the top one only the second one

3. anonymous

The quadratic formula is used to find the roots of a quadratic equation in standard form. For example, given the quadratic equation: $1x ^ {2} + 2x + 2 = 0$ a is the coefficient of x^2, which is 1. b is the coefficient of x, which is 2. c is the constant, which is also 2. The quadratic formula itself is: $(-b \pm \sqrt(b ^ {2} - 4 ac))/ 2a$ You would substitute the values you have from your equation into the formula, and then get two values. Those two values would be your roots for the quadratic equation. $(-(2) \pm \sqrt(2^{2} - 4(1)(2)))/ 2(1)$$= -2 \pm \sqrt(4 - 8) = -2 \pm (\sqrt -4) = -2 \pm 2i$ -2 + 2i would be one solution, -2 - 2i would be the second.

4. anonymous

what does 2i stand for?

5. anonymous

Oh, sorry. The i here stands for imaginary, the square root of negative one. I should have came up with a simpler example. In any case, the presence of the i indicates that there are no real roots for the original quadratic equation. You can use an equation that has real roots (e.g., x^2 + 3x + 2 = 0) with the quadratic formula, and the answer you get will be the same thing as the roots you would get from factoring, if your calculations were not incorrect. (x = -2, -1)