Question

2x^2-x=1

Answer 1

2x^2-x-1=0

Answer 2

\(2x^2-x-1=0\)
\(2x^2-2x+x-1=0\)
\(2x(x-1)+(x-1)=0\)
\((x-1)(2x+1)=0\)
\(x-1=0\) or \(2x+1=0\)
\(x=1\) or \(x=-1/2\)

Answer 3

\[-b \pm \sqrt{b ^{2}-4ac}/2a\]

Answer 4

a=2,b=-1 and c=-1

Answer 5

So right now you have\[2x^2 - x = 1\]

Step 1 : \[2x^2\] = x + 1

Step 2: x (2x - 1) = 1

Step 3 : \[2x^2\] - x - 1 = 0

Do you see that it's simplifying ?

Your answer or solution would be
x = \[\frac{ -1 }{ 2 }\]

Answer X=1

Hope this helped!

Answer 6

I see it now thank you guys so much

Answer 7

it took me a second but I get it now :) @holly01, @kirbykirby

Answer 8

=]

Answer 9

I have a few more if that's alright @kirbykirby

Answer 10

sure