What are the solutions to the equation 36x^2-96x+36=0 ?

The solutions to the equation 36x^2-96x+36=0 are x=(4+sqrt(7))/3 ,x=(4-sqrt(7))/3

Find the zeros of 36x^2-96x+36=0

The zeros of 36x^2-96x+36=0 are x=(4+sqrt(7))/3 ,x=(4-sqrt(7))/3

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36 x^{2} - 96 x + 36 = 0

x = \frac{4 + 7}{3}, x = \frac{4 - 7}{3}

Solution steps

36 x^{2} - 96 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 96 ) \pm ( - 96 ) ^{2} - 4 \cdot 36 \cdot 36}{2 \cdot 36}

(- 96)^{2} - 4 \cdot 36 \cdot 36 = 247

x_{1, 2} = \frac{- ( - 96 ) \pm 24 7}{2 \cdot 36}

Separate the solutions

x_{1} = \frac{- ( - 96 ) + 24 7}{2 \cdot 36}, x_{2} = \frac{- ( - 96 ) - 24 7}{2 \cdot 36}

x = \frac{- ( - 96 ) + 24 7}{2 \cdot 36} : \frac{4 + 7}{3}

x = \frac{- ( - 96 ) - 24 7}{2 \cdot 36} : \frac{4 - 7}{3}

The solutions to the quadratic equation are:

x = \frac{4 + 7}{3}, x = \frac{4 - 7}{3}

2 lo g_{6} (2) + lo g_{6} (9)

(3^{4})^{4}

6 n^{2} - 11 n - 2

4 \cdot 2 \cdot c = 6

\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20}

What are the solutions to the equation 36x^2-96x+36=0 ?
The solutions to the equation 36x^2-96x+36=0 are x=(4+sqrt(7))/3 ,x=(4-sqrt(7))/3
Find the zeros of 36x^2-96x+36=0
The zeros of 36x^2-96x+36=0 are x=(4+sqrt(7))/3 ,x=(4-sqrt(7))/3