What are the solutions to the equation 8x^2-3x-7=0 ?

The solutions to the equation 8x^2-3x-7=0 are x=(3+sqrt(233))/(16),x=(3-sqrt(233))/(16)

Find the zeros of 8x^2-3x-7=0

The zeros of 8x^2-3x-7=0 are x=(3+sqrt(233))/(16),x=(3-sqrt(233))/(16)

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

x = \frac{3 + 233}{16}, x = \frac{3 - 233}{16}

Decimal Notation

x = 1.14152 \dots, x = - 0.76652 \dots

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 8 ( - 7 )}{2 \cdot 8}

(- 3)^{2} - 4 \cdot 8 (- 7) = 233

x_{1, 2} = \frac{- ( - 3 ) \pm 233}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 233}{2 \cdot 8}, x_{2} = \frac{- ( - 3 ) - 233}{2 \cdot 8}

x = \frac{- ( - 3 ) + 233}{2 \cdot 8} : \frac{3 + 233}{16}

x = \frac{- ( - 3 ) - 233}{2 \cdot 8} : \frac{3 - 233}{16}

The solutions to the quadratic equation are:

x = \frac{3 + 233}{16}, x = \frac{3 - 233}{16}

3 k^{2} + 5 k + 1 = 0

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

z^{2} - 9 = 0

5^{2} + 9^{2} = x^{2}

x, x^{2} + 4 x + 1 = 0

What are the solutions to the equation 8x^2-3x-7=0 ?
The solutions to the equation 8x^2-3x-7=0 are x=(3+sqrt(233))/(16),x=(3-sqrt(233))/(16)
Find the zeros of 8x^2-3x-7=0
The zeros of 8x^2-3x-7=0 are x=(3+sqrt(233))/(16),x=(3-sqrt(233))/(16)