What are the solutions to the equation 3=x^2-4x+7 ?

The solutions to the equation 3=x^2-4x+7 are x=2

Find the zeros of 3=x^2-4x+7

The zeros of 3=x^2-4x+7 are x=2

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3 = x^{2} - 4 x + 7

x = 2

Solution steps

3 = x^{2} - 4 x + 7

Switch sides

x^{2} - 4 x + 7 = 3

Move

3

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

4 d^{2} - d - 7 = 2 d^{2}

- 4 x^{2} - 8 x + 1 = 0

x, x^{2} - 6 x - 3 = 0

x (x - 4) = 10

12 x^{2} - 60 x + 50 = 0

What are the solutions to the equation 3=x^2-4x+7 ?
The solutions to the equation 3=x^2-4x+7 are x=2
Find the zeros of 3=x^2-4x+7
The zeros of 3=x^2-4x+7 are x=2