What are the solutions to the equation (x+1)(x+5)=32 ?

The solutions to the equation (x+1)(x+5)=32 are x=3,x=-9

Find the zeros of (x+1)(x+5)=32

The zeros of (x+1)(x+5)=32 are x=3,x=-9

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(x + 1) (x + 5) = 32

x = 3, x = - 9

Solution steps

(x + 1) (x + 5) = 32

Expand

(x + 1) (x + 5) : x^{2} + 6 x + 5

x^{2} + 6 x + 5 = 32

Move

32

to the left side

x^{2} + 6 x - 27 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot ( - 27 )}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot (- 27) = 12

x_{1, 2} = \frac{- 6 \pm 12}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 6 + 12}{2 \cdot 1}, x_{2} = \frac{- 6 - 12}{2 \cdot 1}

x = \frac{- 6 + 12}{2 \cdot 1} : 3

x = \frac{- 6 - 12}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

x = 3, x = - 9

(1.489 - 2.48 I)^{2} + (1.49 - 2.48 I)^{2} = 0

1.25 \cdot 1 0^{- 10} = \frac{x ^{2}}{0.5}

5 x^{2} + 2 x - 39 = 0

h, 0 = \frac{π R ^{2} - 3 π h ^{2}}{4}

(s^{2} + 4) = 0

What are the solutions to the equation (x+1)(x+5)=32 ?
The solutions to the equation (x+1)(x+5)=32 are x=3,x=-9
Find the zeros of (x+1)(x+5)=32
The zeros of (x+1)(x+5)=32 are x=3,x=-9