What are the solutions to the equation z^2-12z+35=0 ?

The solutions to the equation z^2-12z+35=0 are z=7,z=5

Find the zeros of z^2-12z+35=0

The zeros of z^2-12z+35=0 are z=7,z=5

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z^{2} - 12 z + 35 = 0

z = 7, z = 5

Solution steps

z^{2} - 12 z + 35 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 35}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 35 = 2

z_{1, 2} = \frac{- ( - 12 ) \pm 2}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 12 ) + 2}{2 \cdot 1}, z_{2} = \frac{- ( - 12 ) - 2}{2 \cdot 1}

z = \frac{- ( - 12 ) + 2}{2 \cdot 1} : 7

z = \frac{- ( - 12 ) - 2}{2 \cdot 1} : 5

The solutions to the quadratic equation are:

z = 7, z = 5

2 x (2 x + 1) = 5

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

x^{2} + 6 x + 9 = 1

4 a x^{2} - 4 x - (a - 2) = 0

6 x^{2} + 42 x + 60 = 0

What are the solutions to the equation z^2-12z+35=0 ?
The solutions to the equation z^2-12z+35=0 are z=7,z=5
Find the zeros of z^2-12z+35=0
The zeros of z^2-12z+35=0 are z=7,z=5