What are the solutions to the equation m^2+12m=52 ?

The solutions to the equation m^2+12m=52 are m=2(sqrt(22)-3),m=-2(3+sqrt(22))

Find the zeros of m^2+12m=52

The zeros of m^2+12m=52 are m=2(sqrt(22)-3),m=-2(3+sqrt(22))

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m^{2} + 12 m = 52

m = 2 (22 - 3), m = - 2 (3 + 22)

Solution steps

m^{2} + 12 m = 52

Move

52

to the left side

m^{2} + 12 m - 52 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 52 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 52) = 422

m_{1, 2} = \frac{- 12 \pm 4 22}{2 \cdot 1}

Separate the solutions

m_{1} = \frac{- 12 + 4 22}{2 \cdot 1}, m_{2} = \frac{- 12 - 4 22}{2 \cdot 1}

m = \frac{- 12 + 4 22}{2 \cdot 1} : 2 (22 - 3)

m = \frac{- 12 - 4 22}{2 \cdot 1} : - 2 (3 + 22)

The solutions to the quadratic equation are:

m = 2 (22 - 3), m = - 2 (3 + 22)

y^{2} - 5 y = 14

2 x^{2} + 128 x - 4096 = 0

4 (7 - 7 x^{2}) = 7

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

x (x - 7) = 2 x

What are the solutions to the equation m^2+12m=52 ?
The solutions to the equation m^2+12m=52 are m=2(sqrt(22)-3),m=-2(3+sqrt(22))
Find the zeros of m^2+12m=52
The zeros of m^2+12m=52 are m=2(sqrt(22)-3),m=-2(3+sqrt(22))