What are the solutions to the equation z^2+7z-18=0 ?

The solutions to the equation z^2+7z-18=0 are z=2,z=-9

Find the zeros of z^2+7z-18=0

The zeros of z^2+7z-18=0 are z=2,z=-9

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z^{2} + 7 z - 18 = 0

z = 2, z = - 9

Solution steps

z^{2} + 7 z - 18 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 18) = 11

z_{1, 2} = \frac{- 7 \pm 11}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 7 + 11}{2 \cdot 1}, z_{2} = \frac{- 7 - 11}{2 \cdot 1}

z = \frac{- 7 + 11}{2 \cdot 1} : 2

z = \frac{- 7 - 11}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

z = 2, z = - 9

10 - x = 32

x^{4} + 6 x^{2} + 5

x^{2} y^{10} z

9^{2} + 1 2^{2} = c^{2}

0 = 36 x^{2} - 96 x + 48

What are the solutions to the equation z^2+7z-18=0 ?
The solutions to the equation z^2+7z-18=0 are z=2,z=-9
Find the zeros of z^2+7z-18=0
The zeros of z^2+7z-18=0 are z=2,z=-9