What are the solutions to the equation quadraticformula (x-3)(x+9)=-27 ?

The solutions to the equation quadraticformula (x-3)(x+9)=-27 are x=0,x=-6

Find the zeros of quadraticformula (x-3)(x+9)=-27

The zeros of quadraticformula (x-3)(x+9)=-27 are x=0,x=-6

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(x - 3) (x + 9) = - 27

x = 0, x = - 6

Solution steps

(x - 3) (x + 9) = - 27

Expand

(x - 3) (x + 9) : x^{2} + 6 x - 27

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

Move

27

to the left side

x^{2} + 6 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot 0 = 6

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

Separate the solutions

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

x = \frac{- 6 + 6}{2 \cdot 1} : 0

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

The solutions to the quadratic equation are:

x = 0, x = - 6

(- 48) (48)

q, 3 + 5 q = 9 + 4 q

x, 2 x = 14

1. 5^{2}

x, (\frac{1}{4})^{x} = 16

What are the solutions to the equation quadraticformula (x-3)(x+9)=-27 ?
The solutions to the equation quadraticformula (x-3)(x+9)=-27 are x=0,x=-6
Find the zeros of quadraticformula (x-3)(x+9)=-27
The zeros of quadraticformula (x-3)(x+9)=-27 are x=0,x=-6