What are the solutions to the equation 9z^2=-6z+15 ?

The solutions to the equation 9z^2=-6z+15 are z=1,z=-5/3

Find the zeros of 9z^2=-6z+15

The zeros of 9z^2=-6z+15 are z=1,z=-5/3

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9 z^{2} = - 6 z + 15

z = 1, z = - \frac{5}{3}

Solution steps

9 z^{2} = - 6 z + 15

Move

15

to the left side

9 z^{2} - 15 = - 6 z

Move

6 z

to the left side

9 z^{2} - 15 + 6 z = 0

Write in the standard form

a x^{2} + b x + c = 0

9 z^{2} + 6 z - 15 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 9 ( - 15 )}{2 \cdot 9}

6^{2} - 4 \cdot 9 (- 15) = 24

z_{1, 2} = \frac{- 6 \pm 24}{2 \cdot 9}

Separate the solutions

z_{1} = \frac{- 6 + 24}{2 \cdot 9}, z_{2} = \frac{- 6 - 24}{2 \cdot 9}

z = \frac{- 6 + 24}{2 \cdot 9} : 1

z = \frac{- 6 - 24}{2 \cdot 9} : - \frac{5}{3}

The solutions to the quadratic equation are:

z = 1, z = - \frac{5}{3}

3 x^{2} - 27 x = - 5 x - 7

36 a^{4} - 4 a^{2}

1^{0.1}

\frac{7}{27 m ^{2} n} - \frac{8}{18 m ^{4} n ^{6}}

h^{2} - 14 h + 13

What are the solutions to the equation 9z^2=-6z+15 ?
The solutions to the equation 9z^2=-6z+15 are z=1,z=-5/3
Find the zeros of 9z^2=-6z+15
The zeros of 9z^2=-6z+15 are z=1,z=-5/3