What are the solutions to the equation r^2+12r=-32 ?

The solutions to the equation r^2+12r=-32 are r=-4,r=-8

Find the zeros of r^2+12r=-32

The zeros of r^2+12r=-32 are r=-4,r=-8

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r^{2} + 12 r = - 32

r = - 4, r = - 8

Solution steps

r^{2} + 12 r = - 32

Move

32

to the left side

r^{2} + 12 r + 32 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 32 = 4

r_{1, 2} = \frac{- 12 \pm 4}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 12 + 4}{2 \cdot 1}, r_{2} = \frac{- 12 - 4}{2 \cdot 1}

r = \frac{- 12 + 4}{2 \cdot 1} : - 4

r = \frac{- 12 - 4}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

r = - 4, r = - 8

5 x + 4 \leq 3 x^{2}

8 + ∣ 4 x + 1 ∣ = 17

10 x^{2} + 23 x - 42 = 0

(\frac{2}{3} \cdot \frac{4}{5}) + \frac{2}{5} - \frac{3}{10}

25 m^{2} - 81 n^{2}

What are the solutions to the equation r^2+12r=-32 ?
The solutions to the equation r^2+12r=-32 are r=-4,r=-8
Find the zeros of r^2+12r=-32
The zeros of r^2+12r=-32 are r=-4,r=-8