What are the solutions to the equation -3(4x-12)(20+5x)=0 ?

The solutions to the equation -3(4x-12)(20+5x)=0 are x=3,x=-4

Find the zeros of -3(4x-12)(20+5x)=0

The zeros of -3(4x-12)(20+5x)=0 are x=3,x=-4

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- 3 (4 x - 12) (20 + 5 x) = 0

x = 3, x = - 4

Solution steps

- 3 (4 x - 12) (20 + 5 x) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

4 x - 12 = 0 or 20 + 5 x = 0

Solve

4 x - 12 = 0 : x = 3

Solve

20 + 5 x = 0 : x = - 4

The solutions to the quadratic equation are:

x = 3, x = - 4

- 2 (x - 2) (7 x - 2) = 0

x^{2} + 8 x = - 5

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

u^{2} - 9 = (\frac{24}{25}) u^{2}

3 k^{2} - 19 = 15 k - 1

What are the solutions to the equation -3(4x-12)(20+5x)=0 ?
The solutions to the equation -3(4x-12)(20+5x)=0 are x=3,x=-4
Find the zeros of -3(4x-12)(20+5x)=0
The zeros of -3(4x-12)(20+5x)=0 are x=3,x=-4