What are the solutions to the equation m^2+(m-2)^2=340 ?

The solutions to the equation m^2+(m-2)^2=340 are m=14,m=-12

Find the zeros of m^2+(m-2)^2=340

The zeros of m^2+(m-2)^2=340 are m=14,m=-12

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m^{2} + (m - 2)^{2} = 340

m = 14, m = - 12

Solution steps

m^{2} + (m - 2)^{2} = 340

Expand

m^{2} + (m - 2)^{2} : 2 m^{2} - 4 m + 4

2 m^{2} - 4 m + 4 = 340

Move

340

to the left side

2 m^{2} - 4 m - 336 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( - 336 )}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 (- 336) = 52

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

Separate the solutions

m_{1} = \frac{- ( - 4 ) + 52}{2 \cdot 2}, m_{2} = \frac{- ( - 4 ) - 52}{2 \cdot 2}

m = \frac{- ( - 4 ) + 52}{2 \cdot 2} : 14

m = \frac{- ( - 4 ) - 52}{2 \cdot 2} : - 12

The solutions to the quadratic equation are:

m = 14, m = - 12

w^{2} - 6 w + 9

1 0^{- 399.699}

\frac{1}{\frac{5}{1}}

∣ x - 4 ∣ \leq 10

5 x + 10 > 14 x - 8

What are the solutions to the equation m^2+(m-2)^2=340 ?
The solutions to the equation m^2+(m-2)^2=340 are m=14,m=-12
Find the zeros of m^2+(m-2)^2=340
The zeros of m^2+(m-2)^2=340 are m=14,m=-12