What are the solutions to the equation (x+3)(x+5)=48 ?

The solutions to the equation (x+3)(x+5)=48 are x=3,x=-11

Find the zeros of (x+3)(x+5)=48

The zeros of (x+3)(x+5)=48 are x=3,x=-11

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(x + 3) (x + 5) = 48

x = 3, x = - 11

Solution steps

(x + 3) (x + 5) = 48

Expand

(x + 3) (x + 5) : x^{2} + 8 x + 15

x^{2} + 8 x + 15 = 48

Move

48

to the left side

x^{2} + 8 x - 33 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 33 )}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot (- 33) = 14

x_{1, 2} = \frac{- 8 \pm 14}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 8 + 14}{2 \cdot 1}, x_{2} = \frac{- 8 - 14}{2 \cdot 1}

x = \frac{- 8 + 14}{2 \cdot 1} : 3

x = \frac{- 8 - 14}{2 \cdot 1} : - 11

The solutions to the quadratic equation are:

x = 3, x = - 11

4 p^{2} - 4 = 0

x^{2} + 4 x + x = 0

9 v^{2} + 30 v = - 25

7 f^{2} = f + 1

A (x) = x^{2}

What are the solutions to the equation (x+3)(x+5)=48 ?
The solutions to the equation (x+3)(x+5)=48 are x=3,x=-11
Find the zeros of (x+3)(x+5)=48
The zeros of (x+3)(x+5)=48 are x=3,x=-11