What are the solutions to the equation 48r^2+34r-5=0 ?

The solutions to the equation 48r^2+34r-5=0 are r= 1/8 ,r=-5/6

Find the zeros of 48r^2+34r-5=0

The zeros of 48r^2+34r-5=0 are r= 1/8 ,r=-5/6

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48 r^{2} + 34 r - 5 = 0

r = \frac{1}{8}, r = - \frac{5}{6}

Decimal Notation

r = 0.125, r = - 0.83333 \dots

Solution steps

48 r^{2} + 34 r - 5 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 34 \pm 3 4 ^{2} - 4 \cdot 48 ( - 5 )}{2 \cdot 48}

3 4^{2} - 4 \cdot 48 (- 5) = 46

r_{1, 2} = \frac{- 34 \pm 46}{2 \cdot 48}

Separate the solutions

r_{1} = \frac{- 34 + 46}{2 \cdot 48}, r_{2} = \frac{- 34 - 46}{2 \cdot 48}

r = \frac{- 34 + 46}{2 \cdot 48} : \frac{1}{8}

r = \frac{- 34 - 46}{2 \cdot 48} : - \frac{5}{6}

The solutions to the quadratic equation are:

r = \frac{1}{8}, r = - \frac{5}{6}

\frac{e ^{- \frac{1}{29}} \cdot ( \frac{1}{29} ) ^{3}}{3 !}

6 + 9 n \leq 10 n - 9 or 7 + 4 n \leq 2 n - 7

10 ∣ x + 8 ∣ - 13 = - 2 ∣ x + 8 ∣ + 11

y^{2} + 9 y + 20 = 0

What are the solutions to the equation 48r^2+34r-5=0 ?
The solutions to the equation 48r^2+34r-5=0 are r= 1/8 ,r=-5/6
Find the zeros of 48r^2+34r-5=0
The zeros of 48r^2+34r-5=0 are r= 1/8 ,r=-5/6