What are the solutions to the equation 2x^2-x+8=16 ?

The solutions to the equation 2x^2-x+8=16 are x=(1+sqrt(65))/4 ,x=(1-sqrt(65))/4

Find the zeros of 2x^2-x+8=16

The zeros of 2x^2-x+8=16 are x=(1+sqrt(65))/4 ,x=(1-sqrt(65))/4

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2 x^{2} - x + 8 = 16

x = \frac{1 + 65}{4}, x = \frac{1 - 65}{4}

Decimal Notation

x = 2.26556 \dots, x = - 1.76556 \dots

Solution steps

2 x^{2} - x + 8 = 16

Move

16

to the left side

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 2 (- 8) = 65

x_{1, 2} = \frac{- ( - 1 ) \pm 65}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 65}{2 \cdot 2}, x_{2} = \frac{- ( - 1 ) - 65}{2 \cdot 2}

x = \frac{- ( - 1 ) + 65}{2 \cdot 2} : \frac{1 + 65}{4}

x = \frac{- ( - 1 ) - 65}{2 \cdot 2} : \frac{1 - 65}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 65}{4}, x = \frac{1 - 65}{4}

x, y x^{2} - 5 y - x = 0

R, x^{2} + y^{2} = R^{2}

s^{2} + 20 s + 10 = 0

(10 - y)^{2} + y^{2} = 68

x^{2} - 6 x + 9 = 2

What are the solutions to the equation 2x^2-x+8=16 ?
The solutions to the equation 2x^2-x+8=16 are x=(1+sqrt(65))/4 ,x=(1-sqrt(65))/4
Find the zeros of 2x^2-x+8=16
The zeros of 2x^2-x+8=16 are x=(1+sqrt(65))/4 ,x=(1-sqrt(65))/4