What are the solutions to the equation completesquare-4q^2+68q=8 ?

The solutions to the equation completesquare-4q^2+68q=8 are q=(sqrt(281)+17)/2 ,q=(-sqrt(281)+17)/2

Find the zeros of completesquare-4q^2+68q=8

The zeros of completesquare-4q^2+68q=8 are q=(sqrt(281)+17)/2 ,q=(-sqrt(281)+17)/2

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- 4 q^{2} + 68 q = 8

q = \frac{281 + 17}{2}, q = \frac{- 281 + 17}{2}

Decimal Notation

q = 16.88152 \dots, q = 0.11847 \dots

Solution steps

- 4 q^{2} + 68 q = 8

Divide both sides by

- 4

\frac{- 4 q ^{2} + 68 q}{- 4} = \frac{8}{- 4}

q^{2} - 17 q = - 2

Rewrite in the form

(x + a)^{2} = b

(q - \frac{17}{2})^{2} = \frac{281}{4}

For

the solutions are

f (x) = a, - a

Solve

q - \frac{17}{2} = \frac{281}{4} : q = \frac{281 + 17}{2}

Solve

q - \frac{17}{2} = - \frac{281}{4} : q = \frac{- 281 + 17}{2}

The solutions to the quadratic equation are:

q = \frac{281 + 17}{2}, q = \frac{- 281 + 17}{2}

x, m = \frac{y 2 - y 1}{x 2 - x 1}

\frac{5}{x - 1} + \frac{30}{x + 1} = 5

2 + \frac{8}{x - 3} = \frac{4}{x}

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

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

What are the solutions to the equation completesquare-4q^2+68q=8 ?
The solutions to the equation completesquare-4q^2+68q=8 are q=(sqrt(281)+17)/2 ,q=(-sqrt(281)+17)/2
Find the zeros of completesquare-4q^2+68q=8
The zeros of completesquare-4q^2+68q=8 are q=(sqrt(281)+17)/2 ,q=(-sqrt(281)+17)/2