What are the solutions to the equation (x+2)(x+3)=6 ?

The solutions to the equation (x+2)(x+3)=6 are x=0,x=-5

Find the zeros of (x+2)(x+3)=6

The zeros of (x+2)(x+3)=6 are x=0,x=-5

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(x + 2) (x + 3) = 6

x = 0, x = - 5

Solution steps

(x + 2) (x + 3) = 6

Expand

(x + 2) (x + 3) : x^{2} + 5 x + 6

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

Move

6

to the left side

x^{2} + 5 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot 0 = 5

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

Separate the solutions

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

x = \frac{- 5 + 5}{2 \cdot 1} : 0

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

The solutions to the quadratic equation are:

x = 0, x = - 5

4 x^{2} - 12 x - 16 = 0

y^{2} - y = 6

- 7 d - 22 = 4 d

y^{2} - 12 y + 36

x^{2} + 5 x + 13 = 0

What are the solutions to the equation (x+2)(x+3)=6 ?
The solutions to the equation (x+2)(x+3)=6 are x=0,x=-5
Find the zeros of (x+2)(x+3)=6
The zeros of (x+2)(x+3)=6 are x=0,x=-5