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

The solutions to the equation 4x^2-12x+4=0 are x=(3+sqrt(5))/2 ,x=(3-sqrt(5))/2

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

The zeros of 4x^2-12x+4=0 are x=(3+sqrt(5))/2 ,x=(3-sqrt(5))/2

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

x = \frac{3 + 5}{2}, x = \frac{3 - 5}{2}

Decimal Notation

x = 2.61803 \dots, x = 0.38196 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 12)^{2} - 4 \cdot 4 \cdot 4 = 45

x_{1, 2} = \frac{- ( - 12 ) \pm 4 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 4 5}{2 \cdot 4}, x_{2} = \frac{- ( - 12 ) - 4 5}{2 \cdot 4}

x = \frac{- ( - 12 ) + 4 5}{2 \cdot 4} : \frac{3 + 5}{2}

x = \frac{- ( - 12 ) - 4 5}{2 \cdot 4} : \frac{3 - 5}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 5}{2}, x = \frac{3 - 5}{2}

x^{2} - 6 x - 5

- 2 y^{2} = 1 - 3 y^{2}

\frac{1}{2} \cdot 2 \cdot 5^{2} + 2 \cdot 9.8 \cdot 20 = \frac{1}{2} \cdot x^{2}

3 x = x^{2} + 5

(x - 5) (x + 4) = 3 x + 57

What are the solutions to the equation 4x^2-12x+4=0 ?
The solutions to the equation 4x^2-12x+4=0 are x=(3+sqrt(5))/2 ,x=(3-sqrt(5))/2
Find the zeros of 4x^2-12x+4=0
The zeros of 4x^2-12x+4=0 are x=(3+sqrt(5))/2 ,x=(3-sqrt(5))/2