What are the solutions to the equation 8q^2=-2q+1 ?

The solutions to the equation 8q^2=-2q+1 are q= 1/4 ,q=-1/2

Find the zeros of 8q^2=-2q+1

The zeros of 8q^2=-2q+1 are q= 1/4 ,q=-1/2

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8 q^{2} = - 2 q + 1

q = \frac{1}{4}, q = - \frac{1}{2}

Decimal Notation

q = 0.25, q = - 0.5

Solution steps

8 q^{2} = - 2 q + 1

Move

1

to the left side

8 q^{2} - 1 = - 2 q

Move

2 q

to the left side

8 q^{2} - 1 + 2 q = 0

Write in the standard form

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

8 q^{2} + 2 q - 1 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 8 (- 1) = 6

q_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 8}

Separate the solutions

q_{1} = \frac{- 2 + 6}{2 \cdot 8}, q_{2} = \frac{- 2 - 6}{2 \cdot 8}

q = \frac{- 2 + 6}{2 \cdot 8} : \frac{1}{4}

q = \frac{- 2 - 6}{2 \cdot 8} : - \frac{1}{2}

The solutions to the quadratic equation are:

q = \frac{1}{4}, q = - \frac{1}{2}

4 x^{2} - 11 x - 3 = 0

x^{2} - \frac{3}{2} x = 1

- 16 t^{2} + 64 t + 48 = 0

\frac{1}{2} x + 2 = \frac{1}{3} x^{2} + \frac{1}{2}

- x^{2} + 7 x - 3 = 0

What are the solutions to the equation 8q^2=-2q+1 ?
The solutions to the equation 8q^2=-2q+1 are q= 1/4 ,q=-1/2
Find the zeros of 8q^2=-2q+1
The zeros of 8q^2=-2q+1 are q= 1/4 ,q=-1/2