What are the solutions to the equation 5x^2+31x+20=0 ?

The solutions to the equation 5x^2+31x+20=0 are x=(-31+sqrt(561))/(10),x=(-31-sqrt(561))/(10)

Find the zeros of 5x^2+31x+20=0

The zeros of 5x^2+31x+20=0 are x=(-31+sqrt(561))/(10),x=(-31-sqrt(561))/(10)

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5 x^{2} + 31 x + 20 = 0

x = \frac{- 31 + 561}{10}, x = \frac{- 31 - 561}{10}

Decimal Notation

x = - 0.73145 \dots, x = - 5.46854 \dots

Solution steps

5 x^{2} + 31 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 31 \pm 3 1 ^{2} - 4 \cdot 5 \cdot 20}{2 \cdot 5}

3 1^{2} - 4 \cdot 5 \cdot 20 = 561

x_{1, 2} = \frac{- 31 \pm 561}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 31 + 561}{2 \cdot 5}, x_{2} = \frac{- 31 - 561}{2 \cdot 5}

x = \frac{- 31 + 561}{2 \cdot 5} : \frac{- 31 + 561}{10}

x = \frac{- 31 - 561}{2 \cdot 5} : \frac{- 31 - 561}{10}

The solutions to the quadratic equation are:

x = \frac{- 31 + 561}{10}, x = \frac{- 31 - 561}{10}

\frac{2}{3} g - \frac{1}{2} (g - 14) \leq \frac{1}{6} (g + 42)

s^{2} + 4 s + 3

r^{2} + 14 = 0

ln (9.25) = 2.2246

2 (x^{2} + 1) + 1 = 9

What are the solutions to the equation 5x^2+31x+20=0 ?
The solutions to the equation 5x^2+31x+20=0 are x=(-31+sqrt(561))/(10),x=(-31-sqrt(561))/(10)
Find the zeros of 5x^2+31x+20=0
The zeros of 5x^2+31x+20=0 are x=(-31+sqrt(561))/(10),x=(-31-sqrt(561))/(10)