What are the solutions to the equation 4x^2-16x+12=0 ?

The solutions to the equation 4x^2-16x+12=0 are x=3,x=1

Find the zeros of 4x^2-16x+12=0

The zeros of 4x^2-16x+12=0 are x=3,x=1

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4 x^{2} - 16 x + 12 = 0

x = 3, x = 1

Solution steps

4 x^{2} - 16 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 4 \cdot 12}{2 \cdot 4}

(- 16)^{2} - 4 \cdot 4 \cdot 12 = 8

x_{1, 2} = \frac{- ( - 16 ) \pm 8}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 8}{2 \cdot 4}, x_{2} = \frac{- ( - 16 ) - 8}{2 \cdot 4}

x = \frac{- ( - 16 ) + 8}{2 \cdot 4} : 3

x = \frac{- ( - 16 ) - 8}{2 \cdot 4} : 1

The solutions to the quadratic equation are:

x = 3, x = 1

(x - 1)^{2} - 1 = 0

(x + 2) (x + 5) = 28

15 f^{2} + 2 f - 45

\frac{8 k + 9}{18} - \frac{2 k + 1}{18}

∣ x - 7 ∣ = 27

What are the solutions to the equation 4x^2-16x+12=0 ?
The solutions to the equation 4x^2-16x+12=0 are x=3,x=1
Find the zeros of 4x^2-16x+12=0
The zeros of 4x^2-16x+12=0 are x=3,x=1