What are the solutions to the equation 4y^2+12y+9=0 ?

The solutions to the equation 4y^2+12y+9=0 are y=-3/2

Find the zeros of 4y^2+12y+9=0

The zeros of 4y^2+12y+9=0 are y=-3/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 y^{2} + 12 y + 9 = 0

y = - \frac{3}{2}

Solution steps

4 y^{2} + 12 y + 9 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 \cdot 9}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 \cdot 9 = 0

y_{1, 2} = \frac{- 12 \pm 0}{2 \cdot 4}

y = \frac{- 12}{2 \cdot 4}

\frac{- 12}{2 \cdot 4} = - \frac{3}{2}

y = - \frac{3}{2}

The solution to the quadratic equation is:

y = - \frac{3}{2}

5 x - x^{2} = 0

x^{2} + 5^{2} = (52)^{2}

- \frac{1}{4} x + 8 \leq 12

9 x^{6} + 6 x^{3} y + y^{2}

x^{2} + 6 x + 3 = 3

What are the solutions to the equation 4y^2+12y+9=0 ?
The solutions to the equation 4y^2+12y+9=0 are y=-3/2
Find the zeros of 4y^2+12y+9=0
The zeros of 4y^2+12y+9=0 are y=-3/2