What are the solutions to the equation 11x^2-10x+4=6x^2 ?

The solutions to the equation 11x^2-10x+4=6x^2 are x=(5+sqrt(5))/5 ,x=(5-sqrt(5))/5

Find the zeros of 11x^2-10x+4=6x^2

The zeros of 11x^2-10x+4=6x^2 are x=(5+sqrt(5))/5 ,x=(5-sqrt(5))/5

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11 x^{2} - 10 x + 4 = 6 x^{2}

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

Decimal Notation

x = 1.44721 \dots, x = 0.5527864045

Solution steps

11 x^{2} - 10 x + 4 = 6 x^{2}

Move

6 x^{2}

to the left side

5 x^{2} - 10 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 5 \cdot 4}{2 \cdot 5}

(- 10)^{2} - 4 \cdot 5 \cdot 4 = 25

x_{1, 2} = \frac{- ( - 10 ) \pm 2 5}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 2 5}{2 \cdot 5}, x_{2} = \frac{- ( - 10 ) - 2 5}{2 \cdot 5}

x = \frac{- ( - 10 ) + 2 5}{2 \cdot 5} : \frac{5 + 5}{5}

x = \frac{- ( - 10 ) - 2 5}{2 \cdot 5} : \frac{5 - 5}{5}

The solutions to the quadratic equation are:

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

7 - (- 5)

2 = ∣ r - 6 ∣

(\frac{1}{2})^{0.5}

(\frac{p ^{- 3}}{4 p ^{- 6}})^{- 2}

- \frac{x + 5}{2} \leq \frac{12 + 3 x}{4}

What are the solutions to the equation 11x^2-10x+4=6x^2 ?
The solutions to the equation 11x^2-10x+4=6x^2 are x=(5+sqrt(5))/5 ,x=(5-sqrt(5))/5
Find the zeros of 11x^2-10x+4=6x^2
The zeros of 11x^2-10x+4=6x^2 are x=(5+sqrt(5))/5 ,x=(5-sqrt(5))/5