What are the solutions to the equation 2n^2-4=2n ?

The solutions to the equation 2n^2-4=2n are n=2,n=-1

Find the zeros of 2n^2-4=2n

The zeros of 2n^2-4=2n are n=2,n=-1

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2 n^{2} - 4 = 2 n

n = 2, n = - 1

Solution steps

2 n^{2} - 4 = 2 n

Move

2 n

to the left side

2 n^{2} - 4 - 2 n = 0

Write in the standard form

a x^{2} + b x + c = 0

2 n^{2} - 2 n - 4 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 2 (- 4) = 6

n_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 2}, n_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 2}

n = \frac{- ( - 2 ) + 6}{2 \cdot 2} : 2

n = \frac{- ( - 2 ) - 6}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

n = 2, n = - 1

\frac{2}{9} \cdot \frac{6}{5}

8 y^{2} + 30 y + 7

7^{- 2} = \frac{1}{49}

\frac{1 0 ^{9}}{1 0 ^{7}}

a - ab + a^{2}

What are the solutions to the equation 2n^2-4=2n ?
The solutions to the equation 2n^2-4=2n are n=2,n=-1
Find the zeros of 2n^2-4=2n
The zeros of 2n^2-4=2n are n=2,n=-1