What are the solutions to the equation m^2+3m-5=0 ?

The solutions to the equation m^2+3m-5=0 are m=(-3+sqrt(29))/2 ,m=(-3-sqrt(29))/2

Find the zeros of m^2+3m-5=0

The zeros of m^2+3m-5=0 are m=(-3+sqrt(29))/2 ,m=(-3-sqrt(29))/2

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

m = \frac{- 3 + 29}{2}, m = \frac{- 3 - 29}{2}

Decimal Notation

m = 1.19258 \dots, m = - 4.19258 \dots

Solution steps

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

Solve with the quadratic formula

m_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 5) = 29

m_{1, 2} = \frac{- 3 \pm 29}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- 3 + 29}{2 \cdot 1}, m_{2} = \frac{- 3 - 29}{2 \cdot 1}

m = \frac{- 3 + 29}{2 \cdot 1} : \frac{- 3 + 29}{2}

m = \frac{- 3 - 29}{2 \cdot 1} : \frac{- 3 - 29}{2}

The solutions to the quadratic equation are:

m = \frac{- 3 + 29}{2}, m = \frac{- 3 - 29}{2}

m^{2} - 4 m + 6 = 0

3 x^{2} + 56 = x^{2} - 24 x

(2 m - 3) x^{2} + (m - 3) x - 2 m + 1 = 0

p^{2} - 100 = 0

- 28 q = - 96 q^{2} - 2

What are the solutions to the equation m^2+3m-5=0 ?
The solutions to the equation m^2+3m-5=0 are m=(-3+sqrt(29))/2 ,m=(-3-sqrt(29))/2
Find the zeros of m^2+3m-5=0
The zeros of m^2+3m-5=0 are m=(-3+sqrt(29))/2 ,m=(-3-sqrt(29))/2