What are the solutions to the equation r^2+15r+50=6-r ?

The solutions to the equation r^2+15r+50=6-r are r=2(sqrt(5)-4),r=-2(4+sqrt(5))

Find the zeros of r^2+15r+50=6-r

The zeros of r^2+15r+50=6-r are r=2(sqrt(5)-4),r=-2(4+sqrt(5))

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r^{2} + 15 r + 50 = 6 - r

r = 2 (5 - 4), r = - 2 (4 + 5)

Decimal Notation

r = - 3.52786 \dots, r = - 12.47213 \dots

Solution steps

r^{2} + 15 r + 50 = 6 - r

Move

r

to the left side

r^{2} + 16 r + 50 = 6

Move

6

to the left side

r^{2} + 16 r + 44 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot 44}{2 \cdot 1}

1 6^{2} - 4 \cdot 1 \cdot 44 = 45

r_{1, 2} = \frac{- 16 \pm 4 5}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 16 + 4 5}{2 \cdot 1}, r_{2} = \frac{- 16 - 4 5}{2 \cdot 1}

r = \frac{- 16 + 4 5}{2 \cdot 1} : 2 (5 - 4)

r = \frac{- 16 - 4 5}{2 \cdot 1} : - 2 (4 + 5)

The solutions to the quadratic equation are:

r = 2 (5 - 4), r = - 2 (4 + 5)

(p + 1)^{2} = \frac{4}{9}

- 16 t^{2} + 100 = 0

\frac{785}{16} = - 16 (t)^{2} + 50 (t) + 10

32 = - 16 t^{2} + 16 t + 32

s^{2} = 2

What are the solutions to the equation r^2+15r+50=6-r ?
The solutions to the equation r^2+15r+50=6-r are r=2(sqrt(5)-4),r=-2(4+sqrt(5))
Find the zeros of r^2+15r+50=6-r
The zeros of r^2+15r+50=6-r are r=2(sqrt(5)-4),r=-2(4+sqrt(5))