What are the solutions to the equation 8=x^2-4x+3 ?

The solutions to the equation 8=x^2-4x+3 are x=5,x=-1

Find the zeros of 8=x^2-4x+3

The zeros of 8=x^2-4x+3 are x=5,x=-1

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

x = 5, x = - 1

Solution steps

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

Switch sides

x^{2} - 4 x + 3 = 8

Move

8

to the left side

x^{2} - 4 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 5) = 6

x_{1, 2} = \frac{- ( - 4 ) \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 6}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 6}{2 \cdot 1}

x = \frac{- ( - 4 ) + 6}{2 \cdot 1} : 5

x = \frac{- ( - 4 ) - 6}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 5, x = - 1

x^{3} - 5 x^{2} + 6 x - 2

\frac{( x ^{2} + 3 x + 4 )}{4 x - 4}

3 s^{2} + 10 s + 6 = 0

(\frac{1}{32})^{\frac{3}{5}}

- \frac{1}{6} < x \leq 1.8

What are the solutions to the equation 8=x^2-4x+3 ?
The solutions to the equation 8=x^2-4x+3 are x=5,x=-1
Find the zeros of 8=x^2-4x+3
The zeros of 8=x^2-4x+3 are x=5,x=-1