What are the solutions to the equation 49r^2+56r+16=0 ?

The solutions to the equation 49r^2+56r+16=0 are r=-4/7

Find the zeros of 49r^2+56r+16=0

The zeros of 49r^2+56r+16=0 are r=-4/7

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49 r^{2} + 56 r + 16 = 0

r = - \frac{4}{7}

Decimal Notation

r = - 0.57142 \dots

Solution steps

49 r^{2} + 56 r + 16 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 56 \pm 5 6 ^{2} - 4 \cdot 49 \cdot 16}{2 \cdot 49}

5 6^{2} - 4 \cdot 49 \cdot 16 = 0

r_{1, 2} = \frac{- 56 \pm 0}{2 \cdot 49}

r = \frac{- 56}{2 \cdot 49}

\frac{- 56}{2 \cdot 49} = - \frac{4}{7}

r = - \frac{4}{7}

The solution to the quadratic equation is:

r = - \frac{4}{7}

\frac{7}{11} \cdot \frac{6}{10}

\frac{x}{4} - \frac{x}{6} = \frac{1}{4}

x^{4} x^{6}

\frac{2 ( - 0.499 ) + 1}{\frac{1}{4} - ( - 0.499 ) ^{2}}

\frac{1}{6} t^{2} - 4 = 0

What are the solutions to the equation 49r^2+56r+16=0 ?
The solutions to the equation 49r^2+56r+16=0 are r=-4/7
Find the zeros of 49r^2+56r+16=0
The zeros of 49r^2+56r+16=0 are r=-4/7