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

The solutions to the equation 4a^2-8=a are a=(1+sqrt(129))/8 ,a=(1-sqrt(129))/8

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

The zeros of 4a^2-8=a are a=(1+sqrt(129))/8 ,a=(1-sqrt(129))/8

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4 a^{2} - 8 = a

a = \frac{1 + 129}{8}, a = \frac{1 - 129}{8}

Decimal Notation

a = 1.54472 \dots, a = - 1.29472 \dots

Solution steps

4 a^{2} - 8 = a

Move

a

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 4 (- 8) = 129

a_{1, 2} = \frac{- ( - 1 ) \pm 129}{2 \cdot 4}

Separate the solutions

a_{1} = \frac{- ( - 1 ) + 129}{2 \cdot 4}, a_{2} = \frac{- ( - 1 ) - 129}{2 \cdot 4}

a = \frac{- ( - 1 ) + 129}{2 \cdot 4} : \frac{1 + 129}{8}

a = \frac{- ( - 1 ) - 129}{2 \cdot 4} : \frac{1 - 129}{8}

The solutions to the quadratic equation are:

a = \frac{1 + 129}{8}, a = \frac{1 - 129}{8}

4 n - 9 = 2 (5 + 2 n)

∣ x - 5 ∣ = - 1

\frac{π}{\frac{π}{5}}

1 0^{- \frac{1}{2}}

6^{7}

What are the solutions to the equation 4a^2-8=a ?
The solutions to the equation 4a^2-8=a are a=(1+sqrt(129))/8 ,a=(1-sqrt(129))/8
Find the zeros of 4a^2-8=a
The zeros of 4a^2-8=a are a=(1+sqrt(129))/8 ,a=(1-sqrt(129))/8