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

The solutions to the equation 2x^2+5x+7=0 are x=-5/4+i(sqrt(31))/4 ,x=-5/4-i(sqrt(31))/4

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

The zeros of 2x^2+5x+7=0 are x=-5/4+i(sqrt(31))/4 ,x=-5/4-i(sqrt(31))/4

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

x = - \frac{5}{4} + i \frac{31}{4}, x = - \frac{5}{4} - i \frac{31}{4}

Solution steps

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

Solve with the quadratic formula

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

Simplify

5^{2} - 4 \cdot 2 \cdot 7 : 31 i

x_{1, 2} = \frac{- 5 \pm 31 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 31 i}{2 \cdot 2}, x_{2} = \frac{- 5 - 31 i}{2 \cdot 2}

x = \frac{- 5 + 31 i}{2 \cdot 2} : - \frac{5}{4} + i \frac{31}{4}

x = \frac{- 5 - 31 i}{2 \cdot 2} : - \frac{5}{4} - i \frac{31}{4}

The solutions to the quadratic equation are:

x = - \frac{5}{4} + i \frac{31}{4}, x = - \frac{5}{4} - i \frac{31}{4}

16 x - 7 \leq - 7116 x - 7 \leq - 71

(8^{\frac{1}{4}})^{3}

\frac{4 x - 2}{5} + \frac{14 - 3 x}{4} = 3

\frac{8 3}{3} \cdot 2

\frac{9 !}{2 ! \cdot 2 !}

What are the solutions to the equation 2x^2+5x+7=0 ?
The solutions to the equation 2x^2+5x+7=0 are x=-5/4+i(sqrt(31))/4 ,x=-5/4-i(sqrt(31))/4
Find the zeros of 2x^2+5x+7=0
The zeros of 2x^2+5x+7=0 are x=-5/4+i(sqrt(31))/4 ,x=-5/4-i(sqrt(31))/4