What are the solutions to the equation (-5z+1)(-9z-4)=0 ?

The solutions to the equation (-5z+1)(-9z-4)=0 are z= 1/5 ,z=-4/9

Find the zeros of (-5z+1)(-9z-4)=0

The zeros of (-5z+1)(-9z-4)=0 are z= 1/5 ,z=-4/9

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(- 5 z + 1) (- 9 z - 4) = 0

z = \frac{1}{5}, z = - \frac{4}{9}

Decimal Notation

z = 0.2, z = - 0.44444 \dots

Solution steps

(- 5 z + 1) (- 9 z - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

- 5 z + 1 = 0 or - 9 z - 4 = 0

Solve

- 5 z + 1 = 0 : z = \frac{1}{5}

Solve

- 9 z - 4 = 0 : z = - \frac{4}{9}

The solutions to the quadratic equation are:

z = \frac{1}{5}, z = - \frac{4}{9}

t, s = \frac{1}{2} a t^{2}

2 x^{2} - x = 4

25 x^{2} + 60 x + 37 = 0

x, 2 x^{2} = y^{2}

x^{2} - 5 x = 0

What are the solutions to the equation (-5z+1)(-9z-4)=0 ?
The solutions to the equation (-5z+1)(-9z-4)=0 are z= 1/5 ,z=-4/9
Find the zeros of (-5z+1)(-9z-4)=0
The zeros of (-5z+1)(-9z-4)=0 are z= 1/5 ,z=-4/9