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

The solutions to the equation (x+5)(x-4)=3(x+5) are x=-5,x=7

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

The zeros of (x+5)(x-4)=3(x+5) are x=-5,x=7

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(x + 5) (x - 4) = 3 (x + 5)

x = - 5, x = 7

Solution steps

(x + 5) (x - 4) = 3 (x + 5)

Move

3 (x + 5)

to the left side

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

Factor

(x + 5) (x - 4) - 3 (x + 5) : (x + 5) (x - 7)

(x + 5) (x - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 5 = 0 or x - 7 = 0

Solve

x + 5 = 0 : x = - 5

Solve

x - 7 = 0 : x = 7

The solutions to the quadratic equation are:

x = - 5, x = 7

x, y = - \frac{x ^{2}}{2} + \frac{5}{4} x

x, x^{2} + 2 x y = 3 y^{2}

(k + 12)^{2} = 97

5 x^{2} + 9 x - 4 = 3 x + 4

3 - 3 x^{2} = 8 x

What are the solutions to the equation (x+5)(x-4)=3(x+5) ?
The solutions to the equation (x+5)(x-4)=3(x+5) are x=-5,x=7
Find the zeros of (x+5)(x-4)=3(x+5)
The zeros of (x+5)(x-4)=3(x+5) are x=-5,x=7