What are the solutions to the equation 5x^2+3x-7=0 ?

The solutions to the equation 5x^2+3x-7=0 are x=(-3+sqrt(149))/(10),x=(-3-sqrt(149))/(10)

Find the zeros of 5x^2+3x-7=0

The zeros of 5x^2+3x-7=0 are x=(-3+sqrt(149))/(10),x=(-3-sqrt(149))/(10)

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5 x^{2} + 3 x - 7 = 0

x = \frac{- 3 + 149}{10}, x = \frac{- 3 - 149}{10}

Decimal Notation

x = 0.92065 \dots, x = - 1.52065 \dots

Solution steps

5 x^{2} + 3 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 ( - 7 )}{2 \cdot 5}

3^{2} - 4 \cdot 5 (- 7) = 149

x_{1, 2} = \frac{- 3 \pm 149}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 3 + 149}{2 \cdot 5}, x_{2} = \frac{- 3 - 149}{2 \cdot 5}

x = \frac{- 3 + 149}{2 \cdot 5} : \frac{- 3 + 149}{10}

x = \frac{- 3 - 149}{2 \cdot 5} : \frac{- 3 - 149}{10}

The solutions to the quadratic equation are:

x = \frac{- 3 + 149}{10}, x = \frac{- 3 - 149}{10}

- 2 x (x + 1) = 2 (x - 1)

0 = 3 x^{2} - 4 x + 2

- 2 x^{(2)} + 3 x - 5 = 0

x, z = x^{2} + y^{2} - 5 x y - 23 x + 26 y + 9

v, K = \frac{1}{2} m v^{2}

What are the solutions to the equation 5x^2+3x-7=0 ?
The solutions to the equation 5x^2+3x-7=0 are x=(-3+sqrt(149))/(10),x=(-3-sqrt(149))/(10)
Find the zeros of 5x^2+3x-7=0
The zeros of 5x^2+3x-7=0 are x=(-3+sqrt(149))/(10),x=(-3-sqrt(149))/(10)