What are the solutions to the equation 2X^2+4X-5=0 ?

The solutions to the equation 2X^2+4X-5=0 are X=(-2+sqrt(14))/2 ,X=-(2+sqrt(14))/2

Find the zeros of 2X^2+4X-5=0

The zeros of 2X^2+4X-5=0 are X=(-2+sqrt(14))/2 ,X=-(2+sqrt(14))/2

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2 X^{2} + 4 X - 5 = 0

X = \frac{- 2 + 14}{2}, X = - \frac{2 + 14}{2}

Decimal Notation

X = 0.87082 \dots, X = - 2.87082 \dots

Solution steps

2 X^{2} + 4 X - 5 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 2 (- 5) = 214

X_{1, 2} = \frac{- 4 \pm 2 14}{2 \cdot 2}

Separate the solutions

X_{1} = \frac{- 4 + 2 14}{2 \cdot 2}, X_{2} = \frac{- 4 - 2 14}{2 \cdot 2}

X = \frac{- 4 + 2 14}{2 \cdot 2} : \frac{- 2 + 14}{2}

X = \frac{- 4 - 2 14}{2 \cdot 2} : - \frac{2 + 14}{2}

The solutions to the quadratic equation are:

X = \frac{- 2 + 14}{2}, X = - \frac{2 + 14}{2}

(x + 5) (x - 2) (x - 1) (x + 1) < 0

(1 8^{2})^{\frac{1}{3}}

6 x + x (x - 13) = 18

63 n^{2} + 150 n + 75

3 ∣ 2 y + 3 ∣ > 12

What are the solutions to the equation 2X^2+4X-5=0 ?
The solutions to the equation 2X^2+4X-5=0 are X=(-2+sqrt(14))/2 ,X=-(2+sqrt(14))/2
Find the zeros of 2X^2+4X-5=0
The zeros of 2X^2+4X-5=0 are X=(-2+sqrt(14))/2 ,X=-(2+sqrt(14))/2