What are the solutions to the equation completesquare 4x^2-x-8=0 ?

The solutions to the equation completesquare 4x^2-x-8=0 are x=(sqrt(129)+1)/8 ,x=(-sqrt(129)+1)/8

Find the zeros of completesquare 4x^2-x-8=0

The zeros of completesquare 4x^2-x-8=0 are x=(sqrt(129)+1)/8 ,x=(-sqrt(129)+1)/8

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

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

Decimal Notation

x = 1.54472 \dots, x = - 1.29472 \dots

Solution steps

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

Move

8

to the right side

4 x^{2} - x = 8

Divide both sides by

4

\frac{4 x ^{2} - x}{4} = \frac{8}{4}

x^{2} - \frac{x}{4} = 2

Rewrite in the form

(x + a)^{2} = b

(x - \frac{1}{8})^{2} = \frac{129}{64}

For

the solutions are

f (x) = a, - a

Solve

x - \frac{1}{8} = \frac{129}{64} : x = \frac{129 + 1}{8}

Solve

x - \frac{1}{8} = - \frac{129}{64} : x = \frac{- 129 + 1}{8}

The solutions to the quadratic equation are:

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

x^{2} - 2 x + c = 0

(9 x + 3)^{2} = 11

a^{2} + 36 = 13 a

9 x^{2} + 27 x = 0

(3 x + 7)^{2} = 144

What are the solutions to the equation completesquare 4x^2-x-8=0 ?
The solutions to the equation completesquare 4x^2-x-8=0 are x=(sqrt(129)+1)/8 ,x=(-sqrt(129)+1)/8
Find the zeros of completesquare 4x^2-x-8=0
The zeros of completesquare 4x^2-x-8=0 are x=(sqrt(129)+1)/8 ,x=(-sqrt(129)+1)/8