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

The solutions to the equation 4m^2+2m=7 are m=(-1+sqrt(29))/4 ,m=-(1+sqrt(29))/4

Find the zeros of 4m^2+2m=7

The zeros of 4m^2+2m=7 are m=(-1+sqrt(29))/4 ,m=-(1+sqrt(29))/4

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4 m^{2} + 2 m = 7

m = \frac{- 1 + 29}{4}, m = - \frac{1 + 29}{4}

Decimal Notation

m = 1.09629 \dots, m = - 1.59629 \dots

Solution steps

4 m^{2} + 2 m = 7

Move

7

to the left side

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 4 (- 7) = 229

m_{1, 2} = \frac{- 2 \pm 2 29}{2 \cdot 4}

Separate the solutions

m_{1} = \frac{- 2 + 2 29}{2 \cdot 4}, m_{2} = \frac{- 2 - 2 29}{2 \cdot 4}

m = \frac{- 2 + 2 29}{2 \cdot 4} : \frac{- 1 + 29}{4}

m = \frac{- 2 - 2 29}{2 \cdot 4} : - \frac{1 + 29}{4}

The solutions to the quadratic equation are:

m = \frac{- 1 + 29}{4}, m = - \frac{1 + 29}{4}

(\frac{1}{2} x^{3} y^{2} - 4 y)^{9}

2 x^{4} + 7 x^{3} - 87 x^{2} - 395 x - 175

(x - 2)^{2} = 16

1 = x^{2} - 6 x + 9

\frac{5}{2 - i}

What are the solutions to the equation 4m^2+2m=7 ?
The solutions to the equation 4m^2+2m=7 are m=(-1+sqrt(29))/4 ,m=-(1+sqrt(29))/4
Find the zeros of 4m^2+2m=7
The zeros of 4m^2+2m=7 are m=(-1+sqrt(29))/4 ,m=-(1+sqrt(29))/4