What are the solutions to the equation 11a^2-11a+3=6a^2 ?

The solutions to the equation 11a^2-11a+3=6a^2 are a=(11+sqrt(61))/(10),a=(11-sqrt(61))/(10)

Find the zeros of 11a^2-11a+3=6a^2

The zeros of 11a^2-11a+3=6a^2 are a=(11+sqrt(61))/(10),a=(11-sqrt(61))/(10)

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11 a^{2} - 11 a + 3 = 6 a^{2}

a = \frac{11 + 61}{10}, a = \frac{11 - 61}{10}

Decimal Notation

a = 1.88102 \dots, a = 0.31897 \dots

Solution steps

11 a^{2} - 11 a + 3 = 6 a^{2}

Move

6 a^{2}

to the left side

5 a^{2} - 11 a + 3 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 5 \cdot 3}{2 \cdot 5}

(- 11)^{2} - 4 \cdot 5 \cdot 3 = 61

a_{1, 2} = \frac{- ( - 11 ) \pm 61}{2 \cdot 5}

Separate the solutions

a_{1} = \frac{- ( - 11 ) + 61}{2 \cdot 5}, a_{2} = \frac{- ( - 11 ) - 61}{2 \cdot 5}

a = \frac{- ( - 11 ) + 61}{2 \cdot 5} : \frac{11 + 61}{10}

a = \frac{- ( - 11 ) - 61}{2 \cdot 5} : \frac{11 - 61}{10}

The solutions to the quadratic equation are:

a = \frac{11 + 61}{10}, a = \frac{11 - 61}{10}

d^{2} + 11 d + 16 = 0

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

4 b^{2} + 12 b - 3 = 5 b

x, x^{2} = y^{2} + z^{2}

16 t^{2} + 8 t - 8 = - 5

What are the solutions to the equation 11a^2-11a+3=6a^2 ?
The solutions to the equation 11a^2-11a+3=6a^2 are a=(11+sqrt(61))/(10),a=(11-sqrt(61))/(10)
Find the zeros of 11a^2-11a+3=6a^2
The zeros of 11a^2-11a+3=6a^2 are a=(11+sqrt(61))/(10),a=(11-sqrt(61))/(10)