How do you use the quadratic formula to solve for y in $3y^2−9y+4=0$ ?

Question

2935 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-10T17:25:26

y= (9 $+-$ $sqrt33$ )/6

Plug in the values in the formula x= (-b $+-$ $sqrt(b^2-4ac)$ )/2a which represents the solution of a $x^2$ +bx +c=0

y= (9 $+-$ $sqrt(81-4*3*4)$ )/ 2*3

= (9 $+-$ $sqrt33$ )/6

How do you use the quadratic formula to solve for y in 3y^2−9y+4=03y^2−9y+4=0?