## anonymous 3 years ago Can someone show me how I would solve this using the Quadratic Formula? 2x^2 – 16x + 32 = 0

1. anonymous

do you know the quadratic form?

2. anonymous

I think x=4

3. hartnn

Compare your quadratic equation with $$ax^2+bx+c=0$$ find a,b,c then the two roots of x are: $$\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$

4. anonymous

The +- part always confuses me. Could you demonstrate how I'd substitute an equation like this into the form?

5. AccessDenied

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \text{ means:} \\ \quad x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \text{, and} \\ \quad x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} \text{ are the two solutions}$ Basically, its a more compact way of writing both solutions. The typical method is to substitute your values, simplify your sqrt(b^2 - 4ac) and -b, and then deal with splitting the $$\pm$$ into two solutions + and -.

6. anonymous

2x^2-8x-8x+32=0 2x(x-4)-8(x-4)=0 (2x-8)(x-4)=0 x=4,2

7. anonymous

Can I ask why you wrote it as 2x^2-8x-8x+32=0

8. anonymous

its supposed to be b+-4ac?