What are the solutions to the equation 5m^2+7m=7 ?

The solutions to the equation 5m^2+7m=7 are m=(-7+3sqrt(21))/(10),m=(-7-3sqrt(21))/(10)

Find the zeros of 5m^2+7m=7

The zeros of 5m^2+7m=7 are m=(-7+3sqrt(21))/(10),m=(-7-3sqrt(21))/(10)

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5 m^{2} + 7 m = 7

m = \frac{- 7 + 3 21}{10}, m = \frac{- 7 - 3 21}{10}

Decimal Notation

m = 0.67477 \dots, m = - 2.07477 \dots

Solution steps

5 m^{2} + 7 m = 7

Move

7

to the left side

5 m^{2} + 7 m - 7 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 5 ( - 7 )}{2 \cdot 5}

7^{2} - 4 \cdot 5 (- 7) = 321

m_{1, 2} = \frac{- 7 \pm 3 21}{2 \cdot 5}

Separate the solutions

m_{1} = \frac{- 7 + 3 21}{2 \cdot 5}, m_{2} = \frac{- 7 - 3 21}{2 \cdot 5}

m = \frac{- 7 + 3 21}{2 \cdot 5} : \frac{- 7 + 3 21}{10}

m = \frac{- 7 - 3 21}{2 \cdot 5} : \frac{- 7 - 3 21}{10}

The solutions to the quadratic equation are:

m = \frac{- 7 + 3 21}{10}, m = \frac{- 7 - 3 21}{10}

\frac{5 x}{6} - \frac{3 x}{4} = x - 11

x^{2} > 7 x + 18

3 - \frac{1}{3}

6.256 \cdot 1 0^{- 17} = x (1.6 \cdot 1 0^{- 27})

What are the solutions to the equation 5m^2+7m=7 ?
The solutions to the equation 5m^2+7m=7 are m=(-7+3sqrt(21))/(10),m=(-7-3sqrt(21))/(10)
Find the zeros of 5m^2+7m=7
The zeros of 5m^2+7m=7 are m=(-7+3sqrt(21))/(10),m=(-7-3sqrt(21))/(10)