What are the solutions to the equation 5x^2+14x-20=0 ?

The solutions to the equation 5x^2+14x-20=0 are x=(-7+sqrt(149))/5 ,x=-(7+sqrt(149))/5

Find the zeros of 5x^2+14x-20=0

The zeros of 5x^2+14x-20=0 are x=(-7+sqrt(149))/5 ,x=-(7+sqrt(149))/5

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5 x^{2} + 14 x - 20 = 0

x = \frac{- 7 + 149}{5}, x = - \frac{7 + 149}{5}

Decimal Notation

x = 1.04131 \dots, x = - 3.84131 \dots

Solution steps

5 x^{2} + 14 x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 5 ( - 20 )}{2 \cdot 5}

1 4^{2} - 4 \cdot 5 (- 20) = 2149

x_{1, 2} = \frac{- 14 \pm 2 149}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 14 + 2 149}{2 \cdot 5}, x_{2} = \frac{- 14 - 2 149}{2 \cdot 5}

x = \frac{- 14 + 2 149}{2 \cdot 5} : \frac{- 7 + 149}{5}

x = \frac{- 14 - 2 149}{2 \cdot 5} : - \frac{7 + 149}{5}

The solutions to the quadratic equation are:

x = \frac{- 7 + 149}{5}, x = - \frac{7 + 149}{5}

x^{2} + 2 x - 24 > 0

\frac{y ^{- 2} - x ^{- 2}}{y ^{- 2} + x ^{- 2}}

∣ x - 6 ∣ = 2

\frac{7}{x} > 6

(x - 2)^{4} (3 x + 2)^{3} - 10 (x - 2) + 9

What are the solutions to the equation 5x^2+14x-20=0 ?
The solutions to the equation 5x^2+14x-20=0 are x=(-7+sqrt(149))/5 ,x=-(7+sqrt(149))/5
Find the zeros of 5x^2+14x-20=0
The zeros of 5x^2+14x-20=0 are x=(-7+sqrt(149))/5 ,x=-(7+sqrt(149))/5