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

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

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

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

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

x = 2

Solution steps

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

Switch sides

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

Move

x

to the left side

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 4 = 0

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

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

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

x = 2

The solution to the quadratic equation is:

x = 2

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

x^{2} + 6 x - 6

(x - 1) (2 x + 7) = 0

2.4 x^{2} - 3.7 x + 1.4 = 0

6 (x^{2} - 1) - 5 x = 0

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