What are the solutions to the equation 3r^2+5r-1=0 ?

The solutions to the equation 3r^2+5r-1=0 are r=(-5+sqrt(37))/6 ,r=(-5-sqrt(37))/6

Find the zeros of 3r^2+5r-1=0

The zeros of 3r^2+5r-1=0 are r=(-5+sqrt(37))/6 ,r=(-5-sqrt(37))/6

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3 r^{2} + 5 r - 1 = 0

r = \frac{- 5 + 37}{6}, r = \frac{- 5 - 37}{6}

Decimal Notation

r = 0.18046 \dots, r = - 1.84712 \dots

Solution steps

3 r^{2} + 5 r - 1 = 0

Solve with the quadratic formula

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

5^{2} - 4 \cdot 3 (- 1) = 37

r_{1, 2} = \frac{- 5 \pm 37}{2 \cdot 3}

Separate the solutions

r_{1} = \frac{- 5 + 37}{2 \cdot 3}, r_{2} = \frac{- 5 - 37}{2 \cdot 3}

r = \frac{- 5 + 37}{2 \cdot 3} : \frac{- 5 + 37}{6}

r = \frac{- 5 - 37}{2 \cdot 3} : \frac{- 5 - 37}{6}

The solutions to the quadratic equation are:

r = \frac{- 5 + 37}{6}, r = \frac{- 5 - 37}{6}

x^{2} - 400

4 x + 12 > 2

- 14 x + 132 = 26 (x - 18)

\frac{6}{3. 5 ^{- 2}}

5 x - (x - 11) = 21 - 6 x

What are the solutions to the equation 3r^2+5r-1=0 ?
The solutions to the equation 3r^2+5r-1=0 are r=(-5+sqrt(37))/6 ,r=(-5-sqrt(37))/6
Find the zeros of 3r^2+5r-1=0
The zeros of 3r^2+5r-1=0 are r=(-5+sqrt(37))/6 ,r=(-5-sqrt(37))/6