Question

What are some solutions of the equations below? (Quadratic Equation!)

1x^2-3x+1=0

Answer 1

well we need to find what values for x make the left side of the equations equal zero. to do that, you can either factor the equation, or use the quadratic equation.

if there isn't a typo in the equation you gave, then you would have to use the quadratic equation to find the roots

Answer 2

I have the formula written down on a piece of paper...

Answer 3

x = (-b +- sqrt(b^2 - 4ac))/2a

Answer 4

-(b)+ /(6)^2-4(a)(c)

Answer 5

\[x = \frac{ b \pm \sqrt{b^2-4ac}}{ 2a }\]

Answer 6

that's the quadratic formula
where a = 1
b= - 3
and c = 1

Answer 7

Indeed, lol.

Answer 8

sorry, it's suppose to be - b in front... not positive

Answer 9

Yup

Answer 10

Give me a bit more time...