What are the solutions to the equation 2x^2+2=4x ?

The solutions to the equation 2x^2+2=4x are x=1

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2x^2+2=4x

2 x^{2} + 2 = 4 x

x = 1

Solution steps

2 x^{2} + 2 = 4 x

Move

4 x

to the left side

2 x^{2} + 2 - 4 x = 0

Write in the standard form

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

2 x^{2} - 4 x + 2 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 2 \cdot 2 = 0

x_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 2}

x = \frac{- ( - 4 )}{2 \cdot 2}

\frac{- ( - 4 )}{2 \cdot 2} = 1

x = 1

The solution to the quadratic equation is:

x = 1

2 x^{2} = - 10 x + 12

(x - 3) (x + 9) = - 27

(3 x + 2)^{2} = 12 x + 40

3 x^{2} - x = 11

x^{2} - 18 x + 24 = - 7 x

What are the solutions to the equation 2x^2+2=4x ?
The solutions to the equation 2x^2+2=4x are x=1
Find the zeros of 2x^2+2=4x
The zeros of 2x^2+2=4x are x=1