What are the solutions to the equation roots 2x^2+4x+7 ?

The solutions to the equation roots 2x^2+4x+7 are x=-1+isqrt(5/2),x=-1-isqrt(5/2)

Find the zeros of roots 2x^2+4x+7

The zeros of roots 2x^2+4x+7 are x=-1+isqrt(5/2),x=-1-isqrt(5/2)

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roots

2 x^{2} + 4 x + 7

x = - 1 + i \frac{5}{2}, x = - 1 - i \frac{5}{2}

Solution steps

2 x^{2} + 4 x + 7

The roots are the intercepts with the x-axis

(y = 0)

2 x^{2} + 4 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

Simplify

4^{2} - 4 \cdot 2 \cdot 7 : 210 i

x_{1, 2} = \frac{- 4 \pm 2 10 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 2 10 i}{2 \cdot 2}, x_{2} = \frac{- 4 - 2 10 i}{2 \cdot 2}

x = \frac{- 4 + 2 10 i}{2 \cdot 2} : - 1 + i \frac{5}{2}

x = \frac{- 4 - 2 10 i}{2 \cdot 2} : - 1 - i \frac{5}{2}

The solutions to the quadratic equation are:

x = - 1 + i \frac{5}{2}, x = - 1 - i \frac{5}{2}

t, x = v t + \frac{1}{2} a t^{2}

x^{2} = 7^{8}

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

(x + 2)^{2} + 25 = 50

x, x^{2} = 4 p (x) y

What are the solutions to the equation roots 2x^2+4x+7 ?
The solutions to the equation roots 2x^2+4x+7 are x=-1+isqrt(5/2),x=-1-isqrt(5/2)
Find the zeros of roots 2x^2+4x+7
The zeros of roots 2x^2+4x+7 are x=-1+isqrt(5/2),x=-1-isqrt(5/2)