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

The solutions to the equation 2x^2+14x=5 are x=(-7+sqrt(59))/2 ,x=-(7+sqrt(59))/2

Find the zeros of 2x^2+14x=5

The zeros of 2x^2+14x=5 are x=(-7+sqrt(59))/2 ,x=-(7+sqrt(59))/2

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2 x^{2} + 14 x = 5

x = \frac{- 7 + 59}{2}, x = - \frac{7 + 59}{2}

Decimal Notation

x = 0.34057 \dots, x = - 7.34057 \dots

Solution steps

2 x^{2} + 14 x = 5

Move

5

to the left side

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

Solve with the quadratic formula

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

1 4^{2} - 4 \cdot 2 (- 5) = 259

x_{1, 2} = \frac{- 14 \pm 2 59}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 14 + 2 59}{2 \cdot 2}, x_{2} = \frac{- 14 - 2 59}{2 \cdot 2}

x = \frac{- 14 + 2 59}{2 \cdot 2} : \frac{- 7 + 59}{2}

x = \frac{- 14 - 2 59}{2 \cdot 2} : - \frac{7 + 59}{2}

The solutions to the quadratic equation are:

x = \frac{- 7 + 59}{2}, x = - \frac{7 + 59}{2}

\frac{4 - 5 i}{i}

(x + 3)^{2} - 8 x - 9 = 0

7 m - 8 > 6

\frac{x + 2}{3} - \frac{2 x - 1}{2} = 4

\frac{1}{2} x - \frac{4}{3} > 3

What are the solutions to the equation 2x^2+14x=5 ?
The solutions to the equation 2x^2+14x=5 are x=(-7+sqrt(59))/2 ,x=-(7+sqrt(59))/2
Find the zeros of 2x^2+14x=5
The zeros of 2x^2+14x=5 are x=(-7+sqrt(59))/2 ,x=-(7+sqrt(59))/2