What are the solutions to the equation (x+1)(x+2)=6 ?

The solutions to the equation (x+1)(x+2)=6 are x=1,x=-4

Find the zeros of (x+1)(x+2)=6

The zeros of (x+1)(x+2)=6 are x=1,x=-4

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(x + 1) (x + 2) = 6

x = 1, x = - 4

Solution steps

(x + 1) (x + 2) = 6

Expand

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

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

Move

6

to the left side

x^{2} + 3 x - 4 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 4) = 5

x_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 5}{2 \cdot 1}, x_{2} = \frac{- 3 - 5}{2 \cdot 1}

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

x = \frac{- 3 - 5}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 1, x = - 4

10 x^{2} - 250 = 0

x^{2} + 5 = 9

x^{2} + 2 = 6

x^{2} + 3 = 4

a^{2} - 2 a - 48 = 0

What are the solutions to the equation (x+1)(x+2)=6 ?
The solutions to the equation (x+1)(x+2)=6 are x=1,x=-4
Find the zeros of (x+1)(x+2)=6
The zeros of (x+1)(x+2)=6 are x=1,x=-4