What are the solutions to the equation 2x^2-4x+5=3 ?

The solutions to the equation 2x^2-4x+5=3 are x=1

Find the zeros of 2x^2-4x+5=3

The zeros of 2x^2-4x+5=3 are x=1

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2 x^{2} - 4 x + 5 = 3

x = 1

Solution steps

2 x^{2} - 4 x + 5 = 3

Move

3

to the left side

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

30 x - (- x + 6) + (- 5 x + 4) = - (5 x + 6) + (- 8 + 3 x)

\frac{a - \frac{a}{b}}{b - \frac{b}{a}}

2 k^{2} + 3 k - 9

5 w^{2} - 5 w = 0

What are the solutions to the equation 2x^2-4x+5=3 ?
The solutions to the equation 2x^2-4x+5=3 are x=1
Find the zeros of 2x^2-4x+5=3
The zeros of 2x^2-4x+5=3 are x=1