What are the solutions to the equation (x+4)(x+3)=2 ?

The solutions to the equation (x+4)(x+3)=2 are x=-2,x=-5

Find the zeros of (x+4)(x+3)=2

The zeros of (x+4)(x+3)=2 are x=-2,x=-5

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(x + 4) (x + 3) = 2

x = - 2, x = - 5

Solution steps

(x + 4) (x + 3) = 2

Expand

(x + 4) (x + 3) : x^{2} + 7 x + 12

x^{2} + 7 x + 12 = 2

Move

2

to the left side

x^{2} + 7 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 2, x = - 5

- (5)^{- 1}

30 x = 2 (3 x + 8) + 4 (x - 4)

5 x^{2} - 9 x + 4

y^{2} - 10 y = - 18

A^{2} - B^{2} - 12 B - 36

What are the solutions to the equation (x+4)(x+3)=2 ?
The solutions to the equation (x+4)(x+3)=2 are x=-2,x=-5
Find the zeros of (x+4)(x+3)=2
The zeros of (x+4)(x+3)=2 are x=-2,x=-5