What are the solutions to the equation z^2+3z-28=0 ?

The solutions to the equation z^2+3z-28=0 are z=4,z=-7

Find the zeros of z^2+3z-28=0

The zeros of z^2+3z-28=0 are z=4,z=-7

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z^{2} + 3 z - 28 = 0

z = 4, z = - 7

Solution steps

z^{2} + 3 z - 28 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 28) = 11

z_{1, 2} = \frac{- 3 \pm 11}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 3 + 11}{2 \cdot 1}, z_{2} = \frac{- 3 - 11}{2 \cdot 1}

z = \frac{- 3 + 11}{2 \cdot 1} : 4

z = \frac{- 3 - 11}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

z = 4, z = - 7

x (x - 13) \geq - 30

1 0^{0.3}

(2^{3})^{3}

b^{2} - b

(- \frac{8}{17})^{2} + y^{2} = 1

What are the solutions to the equation z^2+3z-28=0 ?
The solutions to the equation z^2+3z-28=0 are z=4,z=-7
Find the zeros of z^2+3z-28=0
The zeros of z^2+3z-28=0 are z=4,z=-7