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

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

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

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

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

x = 6, x = 2

Solution steps

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

x^{2} - 13 x = 30

x^{2} - 6 (x - 1) = 7 (1 - x)

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

x, x (x - c) = a

4 x^{2} + 13 x = 12

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