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 Nov 23, 2018

Best Answer 

 #1
avatar+4330 
+2

We can solve this using quadratic formula, \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) . (Glad that web2.0calc gives you this). Plugging the numbers into the quadratic formula, we have one solution as \(x=\frac{-3+\sqrt{3^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}, \quad \frac{-3+\sqrt{29}}{2}\) and the other solution as \(x=\frac{-3-\sqrt{3^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}:\quad \frac{-3-\sqrt{29}}{2}.\) Thus, the final solutions as \(x=\frac{-3+\sqrt{29}}{2},\:x=\frac{-3-\sqrt{29}}{2}.\)

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 Nov 23, 2018
 #1
avatar+4330 
+2
Best Answer

We can solve this using quadratic formula, \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) . (Glad that web2.0calc gives you this). Plugging the numbers into the quadratic formula, we have one solution as \(x=\frac{-3+\sqrt{3^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}, \quad \frac{-3+\sqrt{29}}{2}\) and the other solution as \(x=\frac{-3-\sqrt{3^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}:\quad \frac{-3-\sqrt{29}}{2}.\) Thus, the final solutions as \(x=\frac{-3+\sqrt{29}}{2},\:x=\frac{-3-\sqrt{29}}{2}.\)

tertre Nov 23, 2018
 #2
avatar+106515 
+2

Excellent, tertre...your LaTex presentations are good...

 

 

cool cool cool

CPhill  Nov 23, 2018
edited by CPhill  Nov 23, 2018
 #3
avatar+4330 
+1

Thank you, CPhill! I have been working on them.

tertre  Nov 23, 2018
 #4
avatar+4116 
+2

umm i'm kinda confused still :(

NotSoSmart  Nov 24, 2018
 #5
avatar+4330 
+2

Sorry if my explanation was confusing. Read more about the quadratic formula https://www.purplemath.com/modules/quadform.htm and https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/a/quadratic-formula-review, and I'm sure that you'll get a deeper understanding. 

tertre  Nov 24, 2018
 #6
avatar+106515 
+1

1x^2 + 3x - 5   = 0

 

We have the form   ax^2 + bx + c   = 0

Where a = 1     b = 3      and    c   = -5

 

The quadraic formula   is

 

[  - b ± √ [  b^2  - (4 * a * c) ]   ]     /    [ 2 * a ]

 

Plugging in for a, b, c......we have

 

[ - 3 ± √ [ 3^2 - (4 * 1 * -5) ] ]  /   [ 2 * 1 ]       simplify

 

[ -3  ± √ [ 9 - ( -20) ] } /  2

 

[ - 3  ± √ 29   ]    /   2

 

And the two solutions as are tertre found

 

x = [ -3  + √ 29   ]    /   2        and       x =    [- 3  - √ 29   ]    /   2

 

 

 

cool cool cool

 Nov 26, 2018

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