What are the solutions to the equation (r-4)(r+8)=-24 ?

The solutions to the equation (r-4)(r+8)=-24 are r=2(sqrt(3)-1),r=-2(1+sqrt(3))

Find the zeros of (r-4)(r+8)=-24

The zeros of (r-4)(r+8)=-24 are r=2(sqrt(3)-1),r=-2(1+sqrt(3))

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(r - 4) (r + 8) = - 24

r = 2 (3 - 1), r = - 2 (1 + 3)

Decimal Notation

r = 1.46410 \dots, r = - 5.46410 \dots

Solution steps

(r - 4) (r + 8) = - 24

Expand

(r - 4) (r + 8) : r^{2} + 4 r - 32

r^{2} + 4 r - 32 = - 24

Move

24

to the left side

r^{2} + 4 r - 8 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 8) = 43

r_{1, 2} = \frac{- 4 \pm 4 3}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 4 + 4 3}{2 \cdot 1}, r_{2} = \frac{- 4 - 4 3}{2 \cdot 1}

r = \frac{- 4 + 4 3}{2 \cdot 1} : 2 (3 - 1)

r = \frac{- 4 - 4 3}{2 \cdot 1} : - 2 (1 + 3)

The solutions to the quadratic equation are:

r = 2 (3 - 1), r = - 2 (1 + 3)

l^{2} - 5 l - 36 = 0

12 x^{2} + 163 x + 15 = 0

1.002 x^{2} - 11.01 x + 0.01265 = 0

5 x^{2} + 3 = 28

(3 x + 6)^{2} - 18 = 0

What are the solutions to the equation (r-4)(r+8)=-24 ?
The solutions to the equation (r-4)(r+8)=-24 are r=2(sqrt(3)-1),r=-2(1+sqrt(3))
Find the zeros of (r-4)(r+8)=-24
The zeros of (r-4)(r+8)=-24 are r=2(sqrt(3)-1),r=-2(1+sqrt(3))