What are the solutions to the equation 40r^2+31r+6=0 ?

The solutions to the equation 40r^2+31r+6=0 are r=-3/8 ,r=-2/5

Find the zeros of 40r^2+31r+6=0

The zeros of 40r^2+31r+6=0 are r=-3/8 ,r=-2/5

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40 r^{2} + 31 r + 6 = 0

r = - \frac{3}{8}, r = - \frac{2}{5}

Decimal Notation

r = - 0.375, r = - 0.4

Solution steps

40 r^{2} + 31 r + 6 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 31 \pm 3 1 ^{2} - 4 \cdot 40 \cdot 6}{2 \cdot 40}

3 1^{2} - 4 \cdot 40 \cdot 6 = 1

r_{1, 2} = \frac{- 31 \pm 1}{2 \cdot 40}

Separate the solutions

r_{1} = \frac{- 31 + 1}{2 \cdot 40}, r_{2} = \frac{- 31 - 1}{2 \cdot 40}

r = \frac{- 31 + 1}{2 \cdot 40} : - \frac{3}{8}

r = \frac{- 31 - 1}{2 \cdot 40} : - \frac{2}{5}

The solutions to the quadratic equation are:

r = - \frac{3}{8}, r = - \frac{2}{5}

F^{2} = 1 0^{2} + 1 0^{2} + 2 \cdot 10 \cdot 10 \cdot \frac{1}{2}

x, x^{2} + y^{2} - 6 x - 6 y + 27 = 0

3 g^{2} - 24 g + 27 = 0

a^{2} - 15 a + 56 = 0

n^{2} - 363 n + 720 = 0

What are the solutions to the equation 40r^2+31r+6=0 ?
The solutions to the equation 40r^2+31r+6=0 are r=-3/8 ,r=-2/5
Find the zeros of 40r^2+31r+6=0
The zeros of 40r^2+31r+6=0 are r=-3/8 ,r=-2/5