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

The solutions to the equation (x+4)(x-12)=5x are x=16,x=-3

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

The zeros of (x+4)(x-12)=5x are x=16,x=-3

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(x + 4) (x - 12) = 5 x

x = 16, x = - 3

Solution steps

(x + 4) (x - 12) = 5 x

Expand

(x + 4) (x - 12) : x^{2} - 8 x - 48

x^{2} - 8 x - 48 = 5 x

Move

5 x

to the left side

x^{2} - 13 x - 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

(- 13)^{2} - 4 \cdot 1 \cdot (- 48) = 19

x_{1, 2} = \frac{- ( - 13 ) \pm 19}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 19}{2 \cdot 1}, x_{2} = \frac{- ( - 13 ) - 19}{2 \cdot 1}

x = \frac{- ( - 13 ) + 19}{2 \cdot 1} : 16

x = \frac{- ( - 13 ) - 19}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 16, x = - 3

4 i^{2}

(x^{2} + 1)^{\frac{3}{2}}

(x - 4) (x + 3) + (x - 4) (x - 3)

3 \cdot 0.6

What are the solutions to the equation (x+4)(x-12)=5x ?
The solutions to the equation (x+4)(x-12)=5x are x=16,x=-3
Find the zeros of (x+4)(x-12)=5x
The zeros of (x+4)(x-12)=5x are x=16,x=-3