What are the solutions to the equation r^2+7r+12=0 ?

The solutions to the equation r^2+7r+12=0 are r=-3,r=-4

Find the zeros of r^2+7r+12=0

The zeros of r^2+7r+12=0 are r=-3,r=-4

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r^{2} + 7 r + 12 = 0

r = - 3, r = - 4

Solution steps

r^{2} + 7 r + 12 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 12 = 1

r_{1, 2} = \frac{- 7 \pm 1}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 7 + 1}{2 \cdot 1}, r_{2} = \frac{- 7 - 1}{2 \cdot 1}

r = \frac{- 7 + 1}{2 \cdot 1} : - 3

r = \frac{- 7 - 1}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

r = - 3, r = - 4

- \frac{1}{32}

\frac{x + 2}{x - 7} \geq \frac{2}{5}

\frac{2 - 3 x}{1 + 2 x} \leq 4

\frac{3 + 4 i}{4 - 2 i}

3 (72 - 32)

What are the solutions to the equation r^2+7r+12=0 ?
The solutions to the equation r^2+7r+12=0 are r=-3,r=-4
Find the zeros of r^2+7r+12=0
The zeros of r^2+7r+12=0 are r=-3,r=-4