What are the solutions to the equation (x+1)(x-5)=16 ?

The solutions to the equation (x+1)(x-5)=16 are x=7,x=-3

Find the zeros of (x+1)(x-5)=16

The zeros of (x+1)(x-5)=16 are x=7,x=-3

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(x + 1) (x - 5) = 16

x = 7, x = - 3

Solution steps

(x + 1) (x - 5) = 16

Expand

(x + 1) (x - 5) : x^{2} - 4 x - 5

x^{2} - 4 x - 5 = 16

Move

16

to the left side

x^{2} - 4 x - 21 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 21 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 21) = 10

x_{1, 2} = \frac{- ( - 4 ) \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 10}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 10}{2 \cdot 1}

x = \frac{- ( - 4 ) + 10}{2 \cdot 1} : 7

x = \frac{- ( - 4 ) - 10}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 7, x = - 3

a^{2} - 8 a + 16 = 0

x^{2} - 34 = 16 x

9 x^{2} - 24 = 1

- 2 (x - 4)^{2} = - 5

6 x^{2} - 10 x - 4 = 0

What are the solutions to the equation (x+1)(x-5)=16 ?
The solutions to the equation (x+1)(x-5)=16 are x=7,x=-3
Find the zeros of (x+1)(x-5)=16
The zeros of (x+1)(x-5)=16 are x=7,x=-3