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formula for quadratic forumula?

Guest May 24, 2017

Best Answer 

 #2
avatar+1493 
+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
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2+0 Answers

 #1
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help

Guest May 24, 2017
 #2
avatar+1493 
+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

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