What are the solutions to the equation (x-4)(x+4)+6(x+4)=3 ?

The solutions to the equation (x-4)(x+4)+6(x+4)=3 are x=-1,x=-5

Find the zeros of (x-4)(x+4)+6(x+4)=3

The zeros of (x-4)(x+4)+6(x+4)=3 are x=-1,x=-5

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(x - 4) (x + 4) + 6 (x + 4) = 3

x = - 1, x = - 5

Solution steps

(x - 4) (x + 4) + 6 (x + 4) = 3

Expand

(x - 4) (x + 4) + 6 (x + 4) : x^{2} + 6 x + 8

x^{2} + 6 x + 8 = 3

Move

3

to the left side

x^{2} + 6 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot 5 = 4

x_{1, 2} = \frac{- 6 \pm 4}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 6 + 4}{2 \cdot 1}, x_{2} = \frac{- 6 - 4}{2 \cdot 1}

x = \frac{- 6 + 4}{2 \cdot 1} : - 1

x = \frac{- 6 - 4}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = - 1, x = - 5

8 p^{2} + 72 = 0

11 x^{2} = - 10 x - 11

m^{2} + 15 m + 50 = 0

9 t^{2} - 15 t - 14 = 0

x^{2} - 4 x - 21

What are the solutions to the equation (x-4)(x+4)+6(x+4)=3 ?
The solutions to the equation (x-4)(x+4)+6(x+4)=3 are x=-1,x=-5
Find the zeros of (x-4)(x+4)+6(x+4)=3
The zeros of (x-4)(x+4)+6(x+4)=3 are x=-1,x=-5