What are the solutions to the equation (x+4)(x+7)=8 ?

The solutions to the equation (x+4)(x+7)=8 are x=(-11+sqrt(41))/2 ,x=(-11-sqrt(41))/2

Find the zeros of (x+4)(x+7)=8

The zeros of (x+4)(x+7)=8 are x=(-11+sqrt(41))/2 ,x=(-11-sqrt(41))/2

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(x + 4) (x + 7) = 8

x = \frac{- 11 + 41}{2}, x = \frac{- 11 - 41}{2}

Decimal Notation

x = - 2.29843 \dots, x = - 8.70156 \dots

Solution steps

(x + 4) (x + 7) = 8

Expand

(x + 4) (x + 7) : x^{2} + 11 x + 28

x^{2} + 11 x + 28 = 8

Move

8

to the left side

x^{2} + 11 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

1 1^{2} - 4 \cdot 1 \cdot 20 = 41

x_{1, 2} = \frac{- 11 \pm 41}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 11 + 41}{2 \cdot 1}, x_{2} = \frac{- 11 - 41}{2 \cdot 1}

x = \frac{- 11 + 41}{2 \cdot 1} : \frac{- 11 + 41}{2}

x = \frac{- 11 - 41}{2 \cdot 1} : \frac{- 11 - 41}{2}

The solutions to the quadratic equation are:

x = \frac{- 11 + 41}{2}, x = \frac{- 11 - 41}{2}

3 x^{2} - 6 x = 12 x

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

4 + 2 x - x^{2}

3 x^{2} - 2 x = 21

2 r^{2} - r - 11 = - 3

What are the solutions to the equation (x+4)(x+7)=8 ?
The solutions to the equation (x+4)(x+7)=8 are x=(-11+sqrt(41))/2 ,x=(-11-sqrt(41))/2
Find the zeros of (x+4)(x+7)=8
The zeros of (x+4)(x+7)=8 are x=(-11+sqrt(41))/2 ,x=(-11-sqrt(41))/2