What are the solutions to the equation 25r^2+80r+64=0 ?

The solutions to the equation 25r^2+80r+64=0 are r=-8/5

Find the zeros of 25r^2+80r+64=0

The zeros of 25r^2+80r+64=0 are r=-8/5

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25 r^{2} + 80 r + 64 = 0

r = - \frac{8}{5}

Decimal Notation

r = - 1.6

Solution steps

25 r^{2} + 80 r + 64 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 \cdot 25 \cdot 64}{2 \cdot 25}

8 0^{2} - 4 \cdot 25 \cdot 64 = 0

r_{1, 2} = \frac{- 80 \pm 0}{2 \cdot 25}

r = \frac{- 80}{2 \cdot 25}

\frac{- 80}{2 \cdot 25} = - \frac{8}{5}

r = - \frac{8}{5}

The solution to the quadratic equation is:

r = - \frac{8}{5}

- x^{2} + 3 x - 2 = 0

x^{2} + 13 x + 12 = 0

\frac{- 7}{3}

x + 6 > 1

y - y^{2} = 0

What are the solutions to the equation 25r^2+80r+64=0 ?
The solutions to the equation 25r^2+80r+64=0 are r=-8/5
Find the zeros of 25r^2+80r+64=0
The zeros of 25r^2+80r+64=0 are r=-8/5