What are the solutions to the equation 4m^2+2m-19=3m^2 ?

The solutions to the equation 4m^2+2m-19=3m^2 are m=-1+2sqrt(5),m=-1-2sqrt(5)

Find the zeros of 4m^2+2m-19=3m^2

The zeros of 4m^2+2m-19=3m^2 are m=-1+2sqrt(5),m=-1-2sqrt(5)

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4 m^{2} + 2 m - 19 = 3 m^{2}

m = - 1 + 25, m = - 1 - 25

Decimal Notation

m = 3.47213 \dots, m = - 5.47213 \dots

Solution steps

4 m^{2} + 2 m - 19 = 3 m^{2}

Move

3 m^{2}

to the left side

m^{2} + 2 m - 19 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 19 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 19) = 45

m_{1, 2} = \frac{- 2 \pm 4 5}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- 2 + 4 5}{2 \cdot 1}, m_{2} = \frac{- 2 - 4 5}{2 \cdot 1}

m = \frac{- 2 + 4 5}{2 \cdot 1} : - 1 + 25

m = \frac{- 2 - 4 5}{2 \cdot 1} : - 1 - 25

The solutions to the quadratic equation are:

m = - 1 + 25, m = - 1 - 25

- 3 x^{2} + 2 x + 5 = 0

2 x^{2} + 8 x + 5

2 x^{2} + 8 x + 3

x^{2} + 44 x = 4.16

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

What are the solutions to the equation 4m^2+2m-19=3m^2 ?
The solutions to the equation 4m^2+2m-19=3m^2 are m=-1+2sqrt(5),m=-1-2sqrt(5)
Find the zeros of 4m^2+2m-19=3m^2
The zeros of 4m^2+2m-19=3m^2 are m=-1+2sqrt(5),m=-1-2sqrt(5)