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

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

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

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

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

x = 2, x = 1

Solution steps

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

Refine

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

Move

3

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = 2, x = 1

\frac{64}{49}

x^{2} + 4 x + 8 = 0

1 \leq \frac{4 x - 5}{- 2} < 9

\frac{3 x - 5}{4} = \frac{x + 1}{2}

1 > \frac{q}{2} - 1

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