What are the solutions to the equation 7(x^2-9)=x(x-3) ?

The solutions to the equation 7(x^2-9)=x(x-3) are x=3,x=-7/2

Find the zeros of 7(x^2-9)=x(x-3)

The zeros of 7(x^2-9)=x(x-3) are x=3,x=-7/2

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7 (x^{2} - 9) = x (x - 3)

x = 3, x = - \frac{7}{2}

Decimal Notation

x = 3, x = - 3.5

Solution steps

7 (x^{2} - 9) = x (x - 3)

Move

x (x - 3)

to the left side

7 (x^{2} - 9) - x (x - 3) = 0

Factor

7 (x^{2} - 9) - x (x - 3) : 3 (x - 3) (2 x + 7)

3 (x - 3) (2 x + 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 3 = 0 or 2 x + 7 = 0

Solve

x - 3 = 0 : x = 3

Solve

2 x + 7 = 0 : x = - \frac{7}{2}

The solutions to the quadratic equation are:

x = 3, x = - \frac{7}{2}

x^{2} - 10 x - 13 = 0

5^{2} + 5^{2} = c^{2}

x^{2} + 6 x + 24 = 0

x^{2} + 26 x = 27

(5 x - 3)^{2} = 64

What are the solutions to the equation 7(x^2-9)=x(x-3) ?
The solutions to the equation 7(x^2-9)=x(x-3) are x=3,x=-7/2
Find the zeros of 7(x^2-9)=x(x-3)
The zeros of 7(x^2-9)=x(x-3) are x=3,x=-7/2