What are the solutions to the equation 9n^2=4+7n ?

The solutions to the equation 9n^2=4+7n are n=(7+sqrt(193))/(18),n=(7-sqrt(193))/(18)

Find the zeros of 9n^2=4+7n

The zeros of 9n^2=4+7n are n=(7+sqrt(193))/(18),n=(7-sqrt(193))/(18)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

9 n^{2} = 4 + 7 n

n = \frac{7 + 193}{18}, n = \frac{7 - 193}{18}

Decimal Notation

n = 1.16069 \dots, n = - 0.38291 \dots

Solution steps

9 n^{2} = 4 + 7 n

Move

7 n

to the left side

9 n^{2} - 7 n = 4

Move

4

to the left side

9 n^{2} - 7 n - 4 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 9 ( - 4 )}{2 \cdot 9}

(- 7)^{2} - 4 \cdot 9 (- 4) = 193

n_{1, 2} = \frac{- ( - 7 ) \pm 193}{2 \cdot 9}

Separate the solutions

n_{1} = \frac{- ( - 7 ) + 193}{2 \cdot 9}, n_{2} = \frac{- ( - 7 ) - 193}{2 \cdot 9}

n = \frac{- ( - 7 ) + 193}{2 \cdot 9} : \frac{7 + 193}{18}

n = \frac{- ( - 7 ) - 193}{2 \cdot 9} : \frac{7 - 193}{18}

The solutions to the quadratic equation are:

n = \frac{7 + 193}{18}, n = \frac{7 - 193}{18}

- 3 - (- 8)

x < 3 x + 40

x (3 x - 18) (6 x - 42) \geq 0

\frac{\frac{1}{2}}{6}

- 7 x \geq 35

What are the solutions to the equation 9n^2=4+7n ?
The solutions to the equation 9n^2=4+7n are n=(7+sqrt(193))/(18),n=(7-sqrt(193))/(18)
Find the zeros of 9n^2=4+7n
The zeros of 9n^2=4+7n are n=(7+sqrt(193))/(18),n=(7-sqrt(193))/(18)