What are the solutions to the equation -12w^2=7+6w ?

The solutions to the equation -12w^2=7+6w are w=-1/4-i(5sqrt(3))/(12),w=-1/4+i(5sqrt(3))/(12)

Find the zeros of -12w^2=7+6w

The zeros of -12w^2=7+6w are w=-1/4-i(5sqrt(3))/(12),w=-1/4+i(5sqrt(3))/(12)

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- 12 w^{2} = 7 + 6 w

w = - \frac{1}{4} - i \frac{5 3}{12}, w = - \frac{1}{4} + i \frac{5 3}{12}

Solution steps

- 12 w^{2} = 7 + 6 w

Move

6 w

to the left side

- 12 w^{2} - 6 w = 7

Move

7

to the left side

- 12 w^{2} - 6 w - 7 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 12 ) ( - 7 )}{2 ( - 12 )}

Simplify

(- 6)^{2} - 4 (- 12) (- 7) : 103 i

w_{1, 2} = \frac{- ( - 6 ) \pm 10 3 i}{2 ( - 12 )}

Separate the solutions

w_{1} = \frac{- ( - 6 ) + 10 3 i}{2 ( - 12 )}, w_{2} = \frac{- ( - 6 ) - 10 3 i}{2 ( - 12 )}

w = \frac{- ( - 6 ) + 10 3 i}{2 ( - 12 )} : - \frac{1}{4} - i \frac{5 3}{12}

w = \frac{- ( - 6 ) - 10 3 i}{2 ( - 12 )} : - \frac{1}{4} + i \frac{5 3}{12}

The solutions to the quadratic equation are:

w = - \frac{1}{4} - i \frac{5 3}{12}, w = - \frac{1}{4} + i \frac{5 3}{12}

96 = 80 t - 16 t^{2}

x^{2} + 4 x + 4 = 0

8^{2} = 5^{2} + x^{2}

2 1^{2} + b^{2} = 2 9^{2}

- 2 x^{2} - 7 x + 15 = 0

What are the solutions to the equation -12w^2=7+6w ?
The solutions to the equation -12w^2=7+6w are w=-1/4-i(5sqrt(3))/(12),w=-1/4+i(5sqrt(3))/(12)
Find the zeros of -12w^2=7+6w
The zeros of -12w^2=7+6w are w=-1/4-i(5sqrt(3))/(12),w=-1/4+i(5sqrt(3))/(12)