What are the solutions to the equation (x+7)(x+5)=3 ?

The solutions to the equation (x+7)(x+5)=3 are x=-4,x=-8

Find the zeros of (x+7)(x+5)=3

The zeros of (x+7)(x+5)=3 are x=-4,x=-8

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(x + 7) (x + 5) = 3

x = - 4, x = - 8

Solution steps

(x + 7) (x + 5) = 3

Expand

(x + 7) (x + 5) : x^{2} + 12 x + 35

x^{2} + 12 x + 35 = 3

Move

3

to the left side

x^{2} + 12 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 32 = 4

x_{1, 2} = \frac{- 12 \pm 4}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 4}{2 \cdot 1}, x_{2} = \frac{- 12 - 4}{2 \cdot 1}

x = \frac{- 12 + 4}{2 \cdot 1} : - 4

x = \frac{- 12 - 4}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = - 4, x = - 8

6 x^{2} - 7 x - 3 = 0

x (x - 6) = - 3

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

x^{2} - 8 x + 15 = 0

n^{2} + 6 n - 27 = 0

What are the solutions to the equation (x+7)(x+5)=3 ?
The solutions to the equation (x+7)(x+5)=3 are x=-4,x=-8
Find the zeros of (x+7)(x+5)=3
The zeros of (x+7)(x+5)=3 are x=-4,x=-8