What are the solutions to the equation (r-2)(r+8)=-5 ?

The solutions to the equation (r-2)(r+8)=-5 are r=-3+2sqrt(5),r=-3-2sqrt(5)

Find the zeros of (r-2)(r+8)=-5

The zeros of (r-2)(r+8)=-5 are r=-3+2sqrt(5),r=-3-2sqrt(5)

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(r - 2) (r + 8) = - 5

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

Decimal Notation

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

Solution steps

(r - 2) (r + 8) = - 5

Expand

(r - 2) (r + 8) : r^{2} + 6 r - 16

r^{2} + 6 r - 16 = - 5

Move

5

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

h^{2} + 5 h - 700 = 0

6 = 2 (x - 3)^{2} - 1

b (x) = x^{2} - 4 x

- 2 y^{2} - 9 = - 9 y - y^{2}

8 = \frac{x}{9} 9 x

What are the solutions to the equation (r-2)(r+8)=-5 ?
The solutions to the equation (r-2)(r+8)=-5 are r=-3+2sqrt(5),r=-3-2sqrt(5)
Find the zeros of (r-2)(r+8)=-5
The zeros of (r-2)(r+8)=-5 are r=-3+2sqrt(5),r=-3-2sqrt(5)