What are the solutions to the equation 8x^2-2x-5=0 ?

The solutions to the equation 8x^2-2x-5=0 are x=(1+sqrt(41))/8 ,x=(1-sqrt(41))/8

Find the zeros of 8x^2-2x-5=0

The zeros of 8x^2-2x-5=0 are x=(1+sqrt(41))/8 ,x=(1-sqrt(41))/8

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8 x^{2} - 2 x - 5 = 0

x = \frac{1 + 41}{8}, x = \frac{1 - 41}{8}

Decimal Notation

x = 0.92539 \dots, x = - 0.67539 \dots

Solution steps

8 x^{2} - 2 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 ( - 5 )}{2 \cdot 8}

(- 2)^{2} - 4 \cdot 8 (- 5) = 241

x_{1, 2} = \frac{- ( - 2 ) \pm 2 41}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 41}{2 \cdot 8}, x_{2} = \frac{- ( - 2 ) - 2 41}{2 \cdot 8}

x = \frac{- ( - 2 ) + 2 41}{2 \cdot 8} : \frac{1 + 41}{8}

x = \frac{- ( - 2 ) - 2 41}{2 \cdot 8} : \frac{1 - 41}{8}

The solutions to the quadratic equation are:

x = \frac{1 + 41}{8}, x = \frac{1 - 41}{8}

0 = x^{2} - 3

2 x^{2} - 9 x + 12 = 0

x^{2} - 10 x + 73 = 0

(x + 2) = (x - 5) (x + 2)

4 x^{2} + 2 = 5 x

What are the solutions to the equation 8x^2-2x-5=0 ?
The solutions to the equation 8x^2-2x-5=0 are x=(1+sqrt(41))/8 ,x=(1-sqrt(41))/8
Find the zeros of 8x^2-2x-5=0
The zeros of 8x^2-2x-5=0 are x=(1+sqrt(41))/8 ,x=(1-sqrt(41))/8