What are the solutions to the equation (x+2)(x+5)=4 ?

The solutions to the equation (x+2)(x+5)=4 are x=-1,x=-6

Find the zeros of (x+2)(x+5)=4

The zeros of (x+2)(x+5)=4 are x=-1,x=-6

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(x + 2) (x + 5) = 4

x = - 1, x = - 6

Solution steps

(x + 2) (x + 5) = 4

Expand

(x + 2) (x + 5) : x^{2} + 7 x + 10

x^{2} + 7 x + 10 = 4

Move

4

to the left side

x^{2} + 7 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 6 = 5

x_{1, 2} = \frac{- 7 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 5}{2 \cdot 1}, x_{2} = \frac{- 7 - 5}{2 \cdot 1}

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

x = \frac{- 7 - 5}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

x = - 1, x = - 6

a, 2 a^{2} b = 2 a^{2} b (1 - 2 a^{3} b^{2})

2 k (k - 5) = 0

- 3 x^{2} + 4 = 0

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

7 x^{2} - 3 x + 5 = 0

What are the solutions to the equation (x+2)(x+5)=4 ?
The solutions to the equation (x+2)(x+5)=4 are x=-1,x=-6
Find the zeros of (x+2)(x+5)=4
The zeros of (x+2)(x+5)=4 are x=-1,x=-6