What are the solutions to the equation 15(2a-2)=5(a^2-1) ?

The solutions to the equation 15(2a-2)=5(a^2-1) are a=1,a=5

Find the zeros of 15(2a-2)=5(a^2-1)

The zeros of 15(2a-2)=5(a^2-1) are a=1,a=5

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15 (2 a - 2) = 5 (a^{2} - 1)

a = 1, a = 5

Solution steps

15 (2 a - 2) = 5 (a^{2} - 1)

Move

5 (a^{2} - 1)

to the left side

15 (2 a - 2) - 5 (a^{2} - 1) = 0

Factor

15 (2 a - 2) - 5 (a^{2} - 1) : - 5 (a - 1) (a - 5)

- 5 (a - 1) (a - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

a - 1 = 0 or a - 5 = 0

Solve

a - 1 = 0 : a = 1

Solve

a - 5 = 0 : a = 5

The solutions to the quadratic equation are:

a = 1, a = 5

2 n^{2} - 14 n + 3 = 3

s^{2} + 3 s + 4

x, x^{2} - 3 m x + 2 m^{2} = 0

- 3 x^{2} = 2 x + 5

s^{2} + 9

What are the solutions to the equation 15(2a-2)=5(a^2-1) ?
The solutions to the equation 15(2a-2)=5(a^2-1) are a=1,a=5
Find the zeros of 15(2a-2)=5(a^2-1)
The zeros of 15(2a-2)=5(a^2-1) are a=1,a=5