What are the solutions to the equation 0=-3y^2+12y-9 ?

The solutions to the equation 0=-3y^2+12y-9 are y=1,y=3

Find the zeros of 0=-3y^2+12y-9

The zeros of 0=-3y^2+12y-9 are y=1,y=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0 = - 3 y^{2} + 12 y - 9

y = 1, y = 3

Solution steps

0 = - 3 y^{2} + 12 y - 9

Switch sides

- 3 y^{2} + 12 y - 9 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 3 ) ( - 9 )}{2 ( - 3 )}

1 2^{2} - 4 (- 3) (- 9) = 6

y_{1, 2} = \frac{- 12 \pm 6}{2 ( - 3 )}

Separate the solutions

y_{1} = \frac{- 12 + 6}{2 ( - 3 )}, y_{2} = \frac{- 12 - 6}{2 ( - 3 )}

y = \frac{- 12 + 6}{2 ( - 3 )} : 1

y = \frac{- 12 - 6}{2 ( - 3 )} : 3

The solutions to the quadratic equation are:

y = 1, y = 3

30 = 10 + 0.9 v - 0.01 v^{2}

z^{2} + z - 72 = 0

x^{2} + 12 x - 16 = 0

p^{2} + 9 p + 20 = 0

4 - y = y^{2} - 4 y + 4

What are the solutions to the equation 0=-3y^2+12y-9 ?
The solutions to the equation 0=-3y^2+12y-9 are y=1,y=3
Find the zeros of 0=-3y^2+12y-9
The zeros of 0=-3y^2+12y-9 are y=1,y=3