What are the solutions to the equation 4x^2-12x+8=0 ?

The solutions to the equation 4x^2-12x+8=0 are x=2,x=1

Find the zeros of 4x^2-12x+8=0

The zeros of 4x^2-12x+8=0 are x=2,x=1

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

x = 2, x = 1

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 4}{2 \cdot 4}, x_{2} = \frac{- ( - 12 ) - 4}{2 \cdot 4}

x = \frac{- ( - 12 ) + 4}{2 \cdot 4} : 2

x = \frac{- ( - 12 ) - 4}{2 \cdot 4} : 1

The solutions to the quadratic equation are:

x = 2, x = 1

8

∣ 4 x - 1 ∣ = 10

- 5 \leq \frac{2 ( x + 3 )}{7} \leq 5

7 k = 3 k - 36

What are the solutions to the equation 4x^2-12x+8=0 ?
The solutions to the equation 4x^2-12x+8=0 are x=2,x=1
Find the zeros of 4x^2-12x+8=0
The zeros of 4x^2-12x+8=0 are x=2,x=1