What are the solutions to the equation 25r^2+15r-4=0 ?

The solutions to the equation 25r^2+15r-4=0 are r= 1/5 ,r=-4/5

Find the zeros of 25r^2+15r-4=0

The zeros of 25r^2+15r-4=0 are r= 1/5 ,r=-4/5

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25 r^{2} + 15 r - 4 = 0

r = \frac{1}{5}, r = - \frac{4}{5}

Solution steps

25 r^{2} + 15 r - 4 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 25 ( - 4 )}{2 \cdot 25}

1 5^{2} - 4 \cdot 25 (- 4) = 25

r_{1, 2} = \frac{- 15 \pm 25}{2 \cdot 25}

Separate the solutions

r_{1} = \frac{- 15 + 25}{2 \cdot 25}, r_{2} = \frac{- 15 - 25}{2 \cdot 25}

r = \frac{- 15 + 25}{2 \cdot 25} : \frac{1}{5}

r = \frac{- 15 - 25}{2 \cdot 25} : - \frac{4}{5}

The solutions to the quadratic equation are:

r = \frac{1}{5}, r = - \frac{4}{5}

m + 7 > 8

- 2 \leq n < 3

\frac{∣ x - 1 ∣}{∣ x - 2 ∣} < 4

\frac{1}{3} \leq (\frac{1}{3})^{x - 1} \leq 9

\frac{1}{2} - \frac{1}{3} x = \frac{4}{3}

What are the solutions to the equation 25r^2+15r-4=0 ?
The solutions to the equation 25r^2+15r-4=0 are r= 1/5 ,r=-4/5
Find the zeros of 25r^2+15r-4=0
The zeros of 25r^2+15r-4=0 are r= 1/5 ,r=-4/5