+0

# help

0
67
2

formula for quadratic forumula?

Guest May 24, 2017

### Best Answer

#2
+815
+1

This is the quadratic formula:

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

Let's try it with an equation to see it in action:

$$x^2+6x-3=0$$

First, find a,b, and c and take note of those values! Remember, a typical quadratic can be written in the form of

$$ax^2+bx+c=0$$. a equals the coefficient of the x^2 term., b equals the coefficient of the x-term, and c equals the constant.

a=1, b=6, c=-3

Now, plug those values into the quadratic formula and solve for x:

$$x = {-6\pm \sqrt{6^2-4(1)(-3)} \over 2(1)}$$

$$x = {-6 \pm \sqrt{36+12)} \over 2}$$

$$x = {-6 \pm \sqrt{48} \over 2}$$

$$x = {-6\pm 4\sqrt{3} \over 2}$$

$$x = -3\pm 2\sqrt{3}$$

Now, try this example on your own:

$$-3x^2-x+14=0$$

TheXSquaredFactor  May 24, 2017
Sort:

### 2+0 Answers

#1
0

help

Guest May 24, 2017
#2
+815
+1
Best Answer

This is the quadratic formula:

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

Let's try it with an equation to see it in action:

$$x^2+6x-3=0$$

First, find a,b, and c and take note of those values! Remember, a typical quadratic can be written in the form of

$$ax^2+bx+c=0$$. a equals the coefficient of the x^2 term., b equals the coefficient of the x-term, and c equals the constant.

a=1, b=6, c=-3

Now, plug those values into the quadratic formula and solve for x:

$$x = {-6\pm \sqrt{6^2-4(1)(-3)} \over 2(1)}$$

$$x = {-6 \pm \sqrt{36+12)} \over 2}$$

$$x = {-6 \pm \sqrt{48} \over 2}$$

$$x = {-6\pm 4\sqrt{3} \over 2}$$

$$x = -3\pm 2\sqrt{3}$$

Now, try this example on your own:

$$-3x^2-x+14=0$$

TheXSquaredFactor  May 24, 2017

### 12 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details