What are the solutions to the equation 1.4x^2+8.3x+1=0 ?

The solutions to the equation 1.4x^2+8.3x+1=0 are x=(-83+sqrt(6329))/(28),x=(-83-sqrt(6329))/(28)

Find the zeros of 1.4x^2+8.3x+1=0

The zeros of 1.4x^2+8.3x+1=0 are x=(-83+sqrt(6329))/(28),x=(-83-sqrt(6329))/(28)

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1.4 x^{2} + 8.3 x + 1 = 0

x = \frac{- 83 + 6329}{28}, x = \frac{- 83 - 6329}{28}

Decimal Notation

x = - 0.12303 \dots, x = - 5.80553 \dots

Solution steps

1.4 x^{2} + 8.3 x + 1 = 0

Multiply both sides by

10

14 x^{2} + 83 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 83 \pm 8 3 ^{2} - 4 \cdot 14 \cdot 10}{2 \cdot 14}

8 3^{2} - 4 \cdot 14 \cdot 10 = 6329

x_{1, 2} = \frac{- 83 \pm 6329}{2 \cdot 14}

Separate the solutions

x_{1} = \frac{- 83 + 6329}{2 \cdot 14}, x_{2} = \frac{- 83 - 6329}{2 \cdot 14}

x = \frac{- 83 + 6329}{2 \cdot 14} : \frac{- 83 + 6329}{28}

x = \frac{- 83 - 6329}{2 \cdot 14} : \frac{- 83 - 6329}{28}

The solutions to the quadratic equation are:

x = \frac{- 83 + 6329}{28}, x = \frac{- 83 - 6329}{28}

3 x^{2} - 7 x + 5 = 1

(x - 2)^{2} = 2 x - 1

x, r^{2} = x^{2} + y^{2}

t, t = \frac{2 a ( b t ^{2} + n )}{3} + \frac{d}{2}

(x - 1)^{2} - 16 = 1

What are the solutions to the equation 1.4x^2+8.3x+1=0 ?
The solutions to the equation 1.4x^2+8.3x+1=0 are x=(-83+sqrt(6329))/(28),x=(-83-sqrt(6329))/(28)
Find the zeros of 1.4x^2+8.3x+1=0
The zeros of 1.4x^2+8.3x+1=0 are x=(-83+sqrt(6329))/(28),x=(-83-sqrt(6329))/(28)