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Solve 6x = 2x^2 + 1

 

 Oct 11, 2020
 #1
avatar+10324 
+1

 

Solve 6x = 2x^2 + 1

 

Hello Guest!

 

\(2x^2-6x+1=0\)

a         b        c 

 

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

\(x = {-3 \pm \sqrt{9-4\cdot 2\cdot 1} \over 2\cdot 1}\)

\(x=\frac{1}{2}\cdot (-3\pm 1)\)

\(x_1=-1\)

\(x_2=-2\)

laugh  !

 Oct 12, 2020
 #2
avatar+27695 
+1

asinus made a small math mistake    b^2 = 36   not   9     ( b is 6    not 3)

using the Quadratic Formula as asinus showed the answer    x =  3/2 +-  sqrt(7) / 2

 Oct 12, 2020
edited by ElectricPavlov  Oct 12, 2020

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