Solve –3x2 + 5x = 1 by using the Quadratic Formula?

3x² - 5x + 1 = 0

x = [- b ± √ (b² - 4ac) ] / 2a

x = [ 5 ± √ (25 - 12) ] / 6

x = [ 5 ± √13 ] / 6

When you have the equation in y = ax² + bx + c =0 form, then:

Quadratic Formula is x= -b/2a ±√(b² -4ac)/2a

This is written as the vertex ± intercepts

Note: When you have x in the equation, use either x^2 or x² for x squared to prevent confusion.

The reason I prefer x= -b/2a ±√(b² -4ac)/2a

is that -b/2a is the vertex, that is, the maximum or minimum value.

We call the term (b2 −4ac) the discriminant. The discriminant is important because it tells you how many roots a quadratic function has. Specifically, if:

1. b² −4ac < 0 There are no real roots, only imaginary.

2. b² −4ac = 0 There is one real root, touches x axis.

3. b² −4ac > 0 There are two real roots.

>>> Simply plug in the numbers to the Quadratic Formula.

(If you hate the Quadratic Formula (QF), or cannot memorize it like me, learn the method of Completing the Squares. This is actually where the QF comes from. I find that method sooooo more appealing. You get the same answers. Watch the link for a video:

-3x^2 + 5x - 1 = 0

x = -[-5 ± √(25 - 12)]/6

= [5 ± √13]/6

You can't. The quadratic formula requires a quadratic equation, which this is not.