What are the solutions to the equation 4y=7y^2-3 ?

The solutions to the equation 4y=7y^2-3 are y=1,y=-3/7

Find the zeros of 4y=7y^2-3

The zeros of 4y=7y^2-3 are y=1,y=-3/7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 y = 7 y^{2} - 3

y = 1, y = - \frac{3}{7}

Decimal Notation

y = 1, y = - 0.42857 \dots

Solution steps

4 y = 7 y^{2} - 3

Switch sides

7 y^{2} - 3 = 4 y

Move

4 y

to the left side

7 y^{2} - 3 - 4 y = 0

Write in the standard form

a x^{2} + b x + c = 0

7 y^{2} - 4 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 7 ( - 3 )}{2 \cdot 7}

(- 4)^{2} - 4 \cdot 7 (- 3) = 10

y_{1, 2} = \frac{- ( - 4 ) \pm 10}{2 \cdot 7}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 10}{2 \cdot 7}, y_{2} = \frac{- ( - 4 ) - 10}{2 \cdot 7}

y = \frac{- ( - 4 ) + 10}{2 \cdot 7} : 1

y = \frac{- ( - 4 ) - 10}{2 \cdot 7} : - \frac{3}{7}

The solutions to the quadratic equation are:

y = 1, y = - \frac{3}{7}

3^{2} + 1^{2} = x^{2}

3 x^{2} - 24 x

(x + 2)^{2} = 50

3 x^{2} - 12 x

x^{2} + 5 x - 126 = 0

What are the solutions to the equation 4y=7y^2-3 ?
The solutions to the equation 4y=7y^2-3 are y=1,y=-3/7
Find the zeros of 4y=7y^2-3
The zeros of 4y=7y^2-3 are y=1,y=-3/7