What are the solutions to the equation (r-3)(r+9)=-16 ?

The solutions to the equation (r-3)(r+9)=-16 are r=-3+2sqrt(5),r=-3-2sqrt(5)

Find the zeros of (r-3)(r+9)=-16

The zeros of (r-3)(r+9)=-16 are r=-3+2sqrt(5),r=-3-2sqrt(5)

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(r - 3) (r + 9) = - 16

r = - 3 + 25, r = - 3 - 25

Decimal Notation

r = 1.47213 \dots, r = - 7.47213 \dots

Solution steps

(r - 3) (r + 9) = - 16

Expand

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

r^{2} + 6 r - 27 = - 16

Move

16

to the left side

r^{2} + 6 r - 11 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot ( - 11 )}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot (- 11) = 45

r_{1, 2} = \frac{- 6 \pm 4 5}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 6 + 4 5}{2 \cdot 1}, r_{2} = \frac{- 6 - 4 5}{2 \cdot 1}

r = \frac{- 6 + 4 5}{2 \cdot 1} : - 3 + 25

r = \frac{- 6 - 4 5}{2 \cdot 1} : - 3 - 25

The solutions to the quadratic equation are:

r = - 3 + 25, r = - 3 - 25

2 x^{2} + 17 x + 15

\frac{7.68 \cdot 1 0 ^{- 4}}{2.4 \cdot 1 0 ^{2}}

\frac{y}{- 4} = - 3

12 5^{\frac{- 4}{3}}

- i \cdot i

What are the solutions to the equation (r-3)(r+9)=-16 ?
The solutions to the equation (r-3)(r+9)=-16 are r=-3+2sqrt(5),r=-3-2sqrt(5)
Find the zeros of (r-3)(r+9)=-16
The zeros of (r-3)(r+9)=-16 are r=-3+2sqrt(5),r=-3-2sqrt(5)