What are the solutions to the equation 2z^2-z-2=0 ?

The solutions to the equation 2z^2-z-2=0 are z=(1+sqrt(17))/4 ,z=(1-sqrt(17))/4

Find the zeros of 2z^2-z-2=0

The zeros of 2z^2-z-2=0 are z=(1+sqrt(17))/4 ,z=(1-sqrt(17))/4

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2 z^{2} - z - 2 = 0

z = \frac{1 + 17}{4}, z = \frac{1 - 17}{4}

Decimal Notation

z = 1.28077 \dots, z = - 0.78077 \dots

Solution steps

2 z^{2} - z - 2 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 2 (- 2) = 17

z_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 \cdot 2}

Separate the solutions

z_{1} = \frac{- ( - 1 ) + 17}{2 \cdot 2}, z_{2} = \frac{- ( - 1 ) - 17}{2 \cdot 2}

z = \frac{- ( - 1 ) + 17}{2 \cdot 2} : \frac{1 + 17}{4}

z = \frac{- ( - 1 ) - 17}{2 \cdot 2} : \frac{1 - 17}{4}

The solutions to the quadratic equation are:

z = \frac{1 + 17}{4}, z = \frac{1 - 17}{4}

x^{2} - 12 x - 30 = 0

x + 3 = x^{2} + x

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

2 x^{2} + 3 x + 1 = 0

x^{2} = (m - 5) x + m^{2}

What are the solutions to the equation 2z^2-z-2=0 ?
The solutions to the equation 2z^2-z-2=0 are z=(1+sqrt(17))/4 ,z=(1-sqrt(17))/4
Find the zeros of 2z^2-z-2=0
The zeros of 2z^2-z-2=0 are z=(1+sqrt(17))/4 ,z=(1-sqrt(17))/4