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

The solutions to the equation 7=9u^2+4u are u=(-2+sqrt(67))/9 ,u=-(2+sqrt(67))/9

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

The zeros of 7=9u^2+4u are u=(-2+sqrt(67))/9 ,u=-(2+sqrt(67))/9

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7 = 9 u^{2} + 4 u

u = \frac{- 2 + 67}{9}, u = - \frac{2 + 67}{9}

Decimal Notation

u = 0.68726 \dots, u = - 1.13170 \dots

Solution steps

7 = 9 u^{2} + 4 u

Switch sides

9 u^{2} + 4 u = 7

Move

7

to the left side

9 u^{2} + 4 u - 7 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 9 (- 7) = 267

u_{1, 2} = \frac{- 4 \pm 2 67}{2 \cdot 9}

Separate the solutions

u_{1} = \frac{- 4 + 2 67}{2 \cdot 9}, u_{2} = \frac{- 4 - 2 67}{2 \cdot 9}

u = \frac{- 4 + 2 67}{2 \cdot 9} : \frac{- 2 + 67}{9}

u = \frac{- 4 - 2 67}{2 \cdot 9} : - \frac{2 + 67}{9}

The solutions to the quadratic equation are:

u = \frac{- 2 + 67}{9}, u = - \frac{2 + 67}{9}

\frac{8 x ^{3} + 22 x ^{2} + 7 x - 10}{2 x ^{2} + 9 x + 9} \leq 0

\frac{( 3 \cdot 5 + 3 )}{10 - 3}

y^{2} + 7 y + 12

- 3 (3 x - 4) = 2 x + 23

(a^{2})^{2}

What are the solutions to the equation 7=9u^2+4u ?
The solutions to the equation 7=9u^2+4u are u=(-2+sqrt(67))/9 ,u=-(2+sqrt(67))/9
Find the zeros of 7=9u^2+4u
The zeros of 7=9u^2+4u are u=(-2+sqrt(67))/9 ,u=-(2+sqrt(67))/9