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

The solutions to the equation (x+1)(x+2)=12 are x=2,x=-5

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

The zeros of (x+1)(x+2)=12 are x=2,x=-5

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(x + 1) (x + 2) = 12

x = 2, x = - 5

Solution steps

(x + 1) (x + 2) = 12

Expand

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

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

Move

12

to the left side

x^{2} + 3 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 10) = 7

x_{1, 2} = \frac{- 3 \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 7}{2 \cdot 1}, x_{2} = \frac{- 3 - 7}{2 \cdot 1}

x = \frac{- 3 + 7}{2 \cdot 1} : 2

x = \frac{- 3 - 7}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 2, x = - 5

(\frac{3}{2} x + 5) \cdot (\frac{7}{2} x - 6) = 30

\frac{1}{4} (4 u + 12) - 1 = 12

∣ 6 - 3 x ∣ - 6 = 9

\frac{x - 1}{x + 2} > 0

x^{2} + 4 x \leq 0

What are the solutions to the equation (x+1)(x+2)=12 ?
The solutions to the equation (x+1)(x+2)=12 are x=2,x=-5
Find the zeros of (x+1)(x+2)=12
The zeros of (x+1)(x+2)=12 are x=2,x=-5