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

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

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

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

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

x = 3, x = 1

Solution steps

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

Move

2

to the left side

x^{2} - 4 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot 3 = 2

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 2}{2 \cdot 1} : 3

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

The solutions to the quadratic equation are:

x = 3, x = 1

\frac{4}{7} + \frac{2}{5}

- 16 k^{2} + 8 k + 24 = 0

\frac{x + 3}{8} \geq 3

0 = 2 x^{2} + 8 x - 24

\frac{x + 10}{3} \geq 2 x

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