What are the solutions to the equation 9r^2+24r+16=0 ?

The solutions to the equation 9r^2+24r+16=0 are r=-4/3

Find the zeros of 9r^2+24r+16=0

The zeros of 9r^2+24r+16=0 are r=-4/3

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9 r^{2} + 24 r + 16 = 0

r = - \frac{4}{3}

Solution steps

9 r^{2} + 24 r + 16 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 9 \cdot 16}{2 \cdot 9}

2 4^{2} - 4 \cdot 9 \cdot 16 = 0

r_{1, 2} = \frac{- 24 \pm 0}{2 \cdot 9}

r = \frac{- 24}{2 \cdot 9}

\frac{- 24}{2 \cdot 9} = - \frac{4}{3}

r = - \frac{4}{3}

The solution to the quadratic equation is:

r = - \frac{4}{3}

y^{2} - 4 y = 2 y - y^{2}

\frac{4 x + 2}{5} \leq 2

∣ 8 y - 4 ∣ > 10

x^{2} + 4 x < 45

7 \cdot \frac{2}{7}

What are the solutions to the equation 9r^2+24r+16=0 ?
The solutions to the equation 9r^2+24r+16=0 are r=-4/3
Find the zeros of 9r^2+24r+16=0
The zeros of 9r^2+24r+16=0 are r=-4/3