What are the solutions to the equation 0=x^2-8x+12 ?

The solutions to the equation 0=x^2-8x+12 are x=6,x=2

Find the zeros of 0=x^2-8x+12

The zeros of 0=x^2-8x+12 are x=6,x=2

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0 = x^{2} - 8 x + 12

x = 6, x = 2

Solution steps

0 = x^{2} - 8 x + 12

Switch sides

x^{2} - 8 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 12 = 4

x_{1, 2} = \frac{- ( - 8 ) \pm 4}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 4}{2 \cdot 1}

x = \frac{- ( - 8 ) + 4}{2 \cdot 1} : 6

x = \frac{- ( - 8 ) - 4}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 6, x = 2

- w^{2} + 5 w - 4 = 0

x^{2} + 2 x = x + 2

4 x^{2} + 24 x + 37 = 0

x^{2} + x + 16 = 0

6 x^{2} = 2 - x

What are the solutions to the equation 0=x^2-8x+12 ?
The solutions to the equation 0=x^2-8x+12 are x=6,x=2
Find the zeros of 0=x^2-8x+12
The zeros of 0=x^2-8x+12 are x=6,x=2