What are the solutions to the equation 8pi=2pi+r(r+3) ?

The solutions to the equation 8pi=2pi+r(r+3) are r=(-3+sqrt(9+24pi))/2 ,r=(-3-sqrt(9+24pi))/2

Find the zeros of 8pi=2pi+r(r+3)

The zeros of 8pi=2pi+r(r+3) are r=(-3+sqrt(9+24pi))/2 ,r=(-3-sqrt(9+24pi))/2

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8 π = 2 π + r (r + 3)

r = \frac{- 3 + 9 + 24 π}{2}, r = \frac{- 3 - 9 + 24 π}{2}

Decimal Notation

r = 3.09342 \dots, r = - 6.09342 \dots

Solution steps

8 π = 2 π + r (r + 3)

Expand

2 π + r (r + 3) : 2 π + r^{2} + 3 r

8 π = 2 π + r^{2} + 3 r

Switch sides

2 π + r^{2} + 3 r = 8 π

Move

8 π

to the left side

r^{2} + 3 r - 6 π = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 6 π )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 6 π) = 9 + 24 π

r_{1, 2} = \frac{- 3 \pm 9 + 24 π}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 3 + 9 + 24 π}{2 \cdot 1}, r_{2} = \frac{- 3 - 9 + 24 π}{2 \cdot 1}

r = \frac{- 3 + 9 + 24 π}{2 \cdot 1} : \frac{- 3 + 9 + 24 π}{2}

r = \frac{- 3 - 9 + 24 π}{2 \cdot 1} : \frac{- 3 - 9 + 24 π}{2}

The solutions to the quadratic equation are:

r = \frac{- 3 + 9 + 24 π}{2}, r = \frac{- 3 - 9 + 24 π}{2}

w^{2} + 8 w + 21 = 0

(x + 4) (x - 3) = - 10

6 c^{2} - 24 = 0

25 - x^{2} = 4 x

C, E = M C^{2}

What are the solutions to the equation 8pi=2pi+r(r+3) ?
The solutions to the equation 8pi=2pi+r(r+3) are r=(-3+sqrt(9+24pi))/2 ,r=(-3-sqrt(9+24pi))/2
Find the zeros of 8pi=2pi+r(r+3)
The zeros of 8pi=2pi+r(r+3) are r=(-3+sqrt(9+24pi))/2 ,r=(-3-sqrt(9+24pi))/2