What are the solutions to the equation 11a^2=-4a+8 ?

The solutions to the equation 11a^2=-4a+8 are a=(2(sqrt(23)-1))/(11),a=-(2(1+sqrt(23)))/(11)

Find the zeros of 11a^2=-4a+8

The zeros of 11a^2=-4a+8 are a=(2(sqrt(23)-1))/(11),a=-(2(1+sqrt(23)))/(11)

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11 a^{2} = - 4 a + 8

a = \frac{2 ( 23 - 1 )}{11}, a = - \frac{2 ( 1 + 23 )}{11}

Decimal Notation

a = 0.69015 \dots, a = - 1.05378 \dots

Solution steps

11 a^{2} = - 4 a + 8

Move

8

to the left side

11 a^{2} - 8 = - 4 a

Move

4 a

to the left side

11 a^{2} - 8 + 4 a = 0

Write in the standard form

a x^{2} + b x + c = 0

11 a^{2} + 4 a - 8 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 11 ( - 8 )}{2 \cdot 11}

4^{2} - 4 \cdot 11 (- 8) = 423

a_{1, 2} = \frac{- 4 \pm 4 23}{2 \cdot 11}

Separate the solutions

a_{1} = \frac{- 4 + 4 23}{2 \cdot 11}, a_{2} = \frac{- 4 - 4 23}{2 \cdot 11}

a = \frac{- 4 + 4 23}{2 \cdot 11} : \frac{2 ( 23 - 1 )}{11}

a = \frac{- 4 - 4 23}{2 \cdot 11} : - \frac{2 ( 1 + 23 )}{11}

The solutions to the quadratic equation are:

a = \frac{2 ( 23 - 1 )}{11}, a = - \frac{2 ( 1 + 23 )}{11}

x^{2} + 110 = 18 x + 45

0 = 2 + 6 t - 4.9 t^{2}

8 x^{2} + 40 x + 50 = 0

x^{2} + 4 x + \frac{7}{4} = 0

6 x^{2} + 3 = 11 x

What are the solutions to the equation 11a^2=-4a+8 ?
The solutions to the equation 11a^2=-4a+8 are a=(2(sqrt(23)-1))/(11),a=-(2(1+sqrt(23)))/(11)
Find the zeros of 11a^2=-4a+8
The zeros of 11a^2=-4a+8 are a=(2(sqrt(23)-1))/(11),a=-(2(1+sqrt(23)))/(11)