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

The solutions to the equation 8x^2-8x=1 are x=(2+sqrt(6))/4 ,x=(2-sqrt(6))/4

Find the zeros of 8x^2-8x=1

The zeros of 8x^2-8x=1 are x=(2+sqrt(6))/4 ,x=(2-sqrt(6))/4

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8 x^{2} - 8 x = 1

x = \frac{2 + 6}{4}, x = \frac{2 - 6}{4}

Decimal Notation

x = 1.11237 \dots, x = - 0.11237 \dots

Solution steps

8 x^{2} - 8 x = 1

Move

1

to the left side

8 x^{2} - 8 x - 1 = 0

Solve with the quadratic formula

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

(- 8)^{2} - 4 \cdot 8 (- 1) = 46

x_{1, 2} = \frac{- ( - 8 ) \pm 4 6}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4 6}{2 \cdot 8}, x_{2} = \frac{- ( - 8 ) - 4 6}{2 \cdot 8}

x = \frac{- ( - 8 ) + 4 6}{2 \cdot 8} : \frac{2 + 6}{4}

x = \frac{- ( - 8 ) - 4 6}{2 \cdot 8} : \frac{2 - 6}{4}

The solutions to the quadratic equation are:

x = \frac{2 + 6}{4}, x = \frac{2 - 6}{4}

5 x^{2} + 40 x - 40 = 0

x, x^{2} + 2 yz + z^{2} = 1

\frac{1}{16} = a^{2}

(2 x - 2)^{2} = 0

s^{2} + 10 s + 5 = 0

What are the solutions to the equation 8x^2-8x=1 ?
The solutions to the equation 8x^2-8x=1 are x=(2+sqrt(6))/4 ,x=(2-sqrt(6))/4
Find the zeros of 8x^2-8x=1
The zeros of 8x^2-8x=1 are x=(2+sqrt(6))/4 ,x=(2-sqrt(6))/4