What are the solutions to the equation -2z^2+58z-80=0 ?

The solutions to the equation -2z^2+58z-80=0 are z=(29-sqrt(681))/2 ,z=(29+sqrt(681))/2

Find the zeros of -2z^2+58z-80=0

The zeros of -2z^2+58z-80=0 are z=(29-sqrt(681))/2 ,z=(29+sqrt(681))/2

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- 2 z^{2} + 58 z - 80 = 0

z = \frac{29 - 681}{2}, z = \frac{29 + 681}{2}

Decimal Notation

z = 1.45201 \dots, z = 27.54798 \dots

Solution steps

- 2 z^{2} + 58 z - 80 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 58 \pm 5 8 ^{2} - 4 ( - 2 ) ( - 80 )}{2 ( - 2 )}

5 8^{2} - 4 (- 2) (- 80) = 2681

z_{1, 2} = \frac{- 58 \pm 2 681}{2 ( - 2 )}

Separate the solutions

z_{1} = \frac{- 58 + 2 681}{2 ( - 2 )}, z_{2} = \frac{- 58 - 2 681}{2 ( - 2 )}

z = \frac{- 58 + 2 681}{2 ( - 2 )} : \frac{29 - 681}{2}

z = \frac{- 58 - 2 681}{2 ( - 2 )} : \frac{29 + 681}{2}

The solutions to the quadratic equation are:

z = \frac{29 - 681}{2}, z = \frac{29 + 681}{2}

- \frac{1}{2} + \frac{1}{4}

\frac{x}{9} = \frac{x}{10} - 14

(7 m) (4 m - 2) = 0

4 x + 5 = x^{2}

- 4 n^{2} + 6 n - 16 = - 5 n^{2}

What are the solutions to the equation -2z^2+58z-80=0 ?
The solutions to the equation -2z^2+58z-80=0 are z=(29-sqrt(681))/2 ,z=(29+sqrt(681))/2
Find the zeros of -2z^2+58z-80=0
The zeros of -2z^2+58z-80=0 are z=(29-sqrt(681))/2 ,z=(29+sqrt(681))/2