What are the solutions to the equation 1+(-2/3)-(-m)m= 9/2 ?

The solutions to the equation 1+(-2/3)-(-m)m= 9/2 are m= 5/(sqrt(6)),m=-5/(sqrt(6))

Find the zeros of 1+(-2/3)-(-m)m= 9/2

The zeros of 1+(-2/3)-(-m)m= 9/2 are m= 5/(sqrt(6)),m=-5/(sqrt(6))

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1 + (- \frac{2}{3}) - (- m) m = \frac{9}{2}

m = \frac{5}{6}, m = - \frac{5}{6}

Decimal Notation

m = 2.04124 \dots, m = - 2.04124 \dots

Solution steps

1 + (- \frac{2}{3}) - (- m) m = \frac{9}{2}

Move

1

to the right side

- \frac{2}{3} - (- m) m = \frac{9}{2} - 1

Move

\frac{2}{3}

to the right side

- (- m) m = \frac{9}{2} + \frac{2}{3} - 1

Simplify

m^{2} = \frac{9}{2} + \frac{2}{3} - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

m = \frac{9}{2} + \frac{2}{3} - 1, m = - \frac{9}{2} + \frac{2}{3} - 1

\frac{9}{2} + \frac{2}{3} - 1 = \frac{5}{6}

- \frac{9}{2} + \frac{2}{3} - 1 = - \frac{5}{6}

m = \frac{5}{6}, m = - \frac{5}{6}

x^{2} = - 56

R, 200 = \frac{1}{3} π R^{2} H

2 x^{2} + 7 x - 184 = 0

8 x^{2} - 23 x - 3 = 0

- 35 x^{2} - 245 x + 630 = 0

What are the solutions to the equation 1+(-2/3)-(-m)m= 9/2 ?
The solutions to the equation 1+(-2/3)-(-m)m= 9/2 are m= 5/(sqrt(6)),m=-5/(sqrt(6))
Find the zeros of 1+(-2/3)-(-m)m= 9/2
The zeros of 1+(-2/3)-(-m)m= 9/2 are m= 5/(sqrt(6)),m=-5/(sqrt(6))