What are the solutions to the equation m^2+12m=3 ?

The solutions to the equation m^2+12m=3 are m=-6+sqrt(39),m=-6-sqrt(39)

Find the zeros of m^2+12m=3

The zeros of m^2+12m=3 are m=-6+sqrt(39),m=-6-sqrt(39)

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m^{2} + 12 m = 3

m = - 6 + 39, m = - 6 - 39

Decimal Notation

m = 0.24499 \dots, m = - 12.24499 \dots

Solution steps

m^{2} + 12 m = 3

Move

3

to the left side

m^{2} + 12 m - 3 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 3) = 239

m_{1, 2} = \frac{- 12 \pm 2 39}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- 12 + 2 39}{2 \cdot 1}, m_{2} = \frac{- 12 - 2 39}{2 \cdot 1}

m = \frac{- 12 + 2 39}{2 \cdot 1} : - 6 + 39

m = \frac{- 12 - 2 39}{2 \cdot 1} : - 6 - 39

The solutions to the quadratic equation are:

m = - 6 + 39, m = - 6 - 39

- 0.002 x^{2} - 0.4 x + 230 = 0.0004 x^{2} + 0.1 x + 156

(x - 7) (x + 9) = 0

x, x^{2} + 23 x y - y^{2} + 2 = 0

x^{2} = - 10 x - 1

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

What are the solutions to the equation m^2+12m=3 ?
The solutions to the equation m^2+12m=3 are m=-6+sqrt(39),m=-6-sqrt(39)
Find the zeros of m^2+12m=3
The zeros of m^2+12m=3 are m=-6+sqrt(39),m=-6-sqrt(39)