What are the solutions to the equation 3y^2-36y+81=0 ?

The solutions to the equation 3y^2-36y+81=0 are y=9,y=3

Find the zeros of 3y^2-36y+81=0

The zeros of 3y^2-36y+81=0 are y=9,y=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 y^{2} - 36 y + 81 = 0

y = 9, y = 3

Solution steps

3 y^{2} - 36 y + 81 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 3 \cdot 81}{2 \cdot 3}

(- 36)^{2} - 4 \cdot 3 \cdot 81 = 18

y_{1, 2} = \frac{- ( - 36 ) \pm 18}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 36 ) + 18}{2 \cdot 3}, y_{2} = \frac{- ( - 36 ) - 18}{2 \cdot 3}

y = \frac{- ( - 36 ) + 18}{2 \cdot 3} : 9

y = \frac{- ( - 36 ) - 18}{2 \cdot 3} : 3

The solutions to the quadratic equation are:

y = 9, y = 3

4 x^{2} = - 3 (2 x)

w^{2} - 4 w - 5 = 0

s^{2} + 5 s + 1 = 0

y - \frac{y ^{2}}{2} = 0

a^{2} + 6^{2} = 9^{2}

What are the solutions to the equation 3y^2-36y+81=0 ?
The solutions to the equation 3y^2-36y+81=0 are y=9,y=3
Find the zeros of 3y^2-36y+81=0
The zeros of 3y^2-36y+81=0 are y=9,y=3