What are the solutions to the equation ((x+3)(x+1))/2 =12 ?

The solutions to the equation ((x+3)(x+1))/2 =12 are x=3,x=-7

Find the zeros of ((x+3)(x+1))/2 =12

The zeros of ((x+3)(x+1))/2 =12 are x=3,x=-7

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\frac{( x + 3 ) ( x + 1 )}{2} = 12

x = 3, x = - 7

Solution steps

\frac{( x + 3 ) ( x + 1 )}{2} = 12

Multiply both sides by

2

(x + 3) (x + 1) = 24

Expand

(x + 3) (x + 1) : x^{2} + 4 x + 3

x^{2} + 4 x + 3 = 24

Move

24

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} : 3

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

The solutions to the quadratic equation are:

x = 3, x = - 7

y^{2} = - 7 y

y^{2} = - 28

2 p^{2} + 8 p + (k + 1) = 0

3 x^{2} - 11 x - 42 = 0

n^{2} + 2 n - 120 = 0

What are the solutions to the equation ((x+3)(x+1))/2 =12 ?
The solutions to the equation ((x+3)(x+1))/2 =12 are x=3,x=-7
Find the zeros of ((x+3)(x+1))/2 =12
The zeros of ((x+3)(x+1))/2 =12 are x=3,x=-7