What are the solutions to the equation 2z^2+5z=1 ?

The solutions to the equation 2z^2+5z=1 are z=(-5+sqrt(33))/4 ,z=(-5-sqrt(33))/4

Find the zeros of 2z^2+5z=1

The zeros of 2z^2+5z=1 are z=(-5+sqrt(33))/4 ,z=(-5-sqrt(33))/4

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2 z^{2} + 5 z = 1

z = \frac{- 5 + 33}{4}, z = \frac{- 5 - 33}{4}

Decimal Notation

z = 0.18614 \dots, z = - 2.68614 \dots

Solution steps

2 z^{2} + 5 z = 1

Move

1

to the left side

2 z^{2} + 5 z - 1 = 0

Solve with the quadratic formula

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

5^{2} - 4 \cdot 2 (- 1) = 33

z_{1, 2} = \frac{- 5 \pm 33}{2 \cdot 2}

Separate the solutions

z_{1} = \frac{- 5 + 33}{2 \cdot 2}, z_{2} = \frac{- 5 - 33}{2 \cdot 2}

z = \frac{- 5 + 33}{2 \cdot 2} : \frac{- 5 + 33}{4}

z = \frac{- 5 - 33}{2 \cdot 2} : \frac{- 5 - 33}{4}

The solutions to the quadratic equation are:

z = \frac{- 5 + 33}{4}, z = \frac{- 5 - 33}{4}

t, f = (t - 3) u^{2} (t) - (t - 2) u^{3} (t)

2 x^{2} = - 32

8 x^{2} - 15 x - 1 = - 7 x

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

- 2 x^{2} + 12 x - 14 = 0

What are the solutions to the equation 2z^2+5z=1 ?
The solutions to the equation 2z^2+5z=1 are z=(-5+sqrt(33))/4 ,z=(-5-sqrt(33))/4
Find the zeros of 2z^2+5z=1
The zeros of 2z^2+5z=1 are z=(-5+sqrt(33))/4 ,z=(-5-sqrt(33))/4