What are the solutions to the equation 2x^2+3/7 x-8=0 ?

The solutions to the equation 2x^2+3/7 x-8=0 are x=(-3+sqrt(3145))/(28),x=(-3-sqrt(3145))/(28)

Find the zeros of 2x^2+3/7 x-8=0

The zeros of 2x^2+3/7 x-8=0 are x=(-3+sqrt(3145))/(28),x=(-3-sqrt(3145))/(28)

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2 x^{2} + \frac{3}{7} x - 8 = 0

x = \frac{- 3 + 3145}{28}, x = \frac{- 3 - 3145}{28}

Decimal Notation

x = 1.89572 \dots, x = - 2.11001 \dots

Solution steps

2 x^{2} + \frac{3}{7} x - 8 = 0

Multiply both sides by

7

14 x^{2} + 3 x - 56 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 14 ( - 56 )}{2 \cdot 14}

3^{2} - 4 \cdot 14 (- 56) = 3145

x_{1, 2} = \frac{- 3 \pm 3145}{2 \cdot 14}

Separate the solutions

x_{1} = \frac{- 3 + 3145}{2 \cdot 14}, x_{2} = \frac{- 3 - 3145}{2 \cdot 14}

x = \frac{- 3 + 3145}{2 \cdot 14} : \frac{- 3 + 3145}{28}

x = \frac{- 3 - 3145}{2 \cdot 14} : \frac{- 3 - 3145}{28}

The solutions to the quadratic equation are:

x = \frac{- 3 + 3145}{28}, x = \frac{- 3 - 3145}{28}

(x - 4)^{2} - 121 = 0

x^{2} = 12 x

x, z = 4 - x^{2} - y^{2}

y, x^{2} + x y^{2} + y^{2} = 1

m^{2} + m + 6 = 0

What are the solutions to the equation 2x^2+3/7 x-8=0 ?
The solutions to the equation 2x^2+3/7 x-8=0 are x=(-3+sqrt(3145))/(28),x=(-3-sqrt(3145))/(28)
Find the zeros of 2x^2+3/7 x-8=0
The zeros of 2x^2+3/7 x-8=0 are x=(-3+sqrt(3145))/(28),x=(-3-sqrt(3145))/(28)