What are the solutions to the equation 2x^2+7x+15=0 ?

The solutions to the equation 2x^2+7x+15=0 are x=-7/4+i(sqrt(71))/4 ,x=-7/4-i(sqrt(71))/4

Find the zeros of 2x^2+7x+15=0

The zeros of 2x^2+7x+15=0 are x=-7/4+i(sqrt(71))/4 ,x=-7/4-i(sqrt(71))/4

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2 x^{2} + 7 x + 15 = 0

x = - \frac{7}{4} + i \frac{71}{4}, x = - \frac{7}{4} - i \frac{71}{4}

Solution steps

2 x^{2} + 7 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 \cdot 15}{2 \cdot 2}

Simplify

7^{2} - 4 \cdot 2 \cdot 15 : 71 i

x_{1, 2} = \frac{- 7 \pm 71 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 71 i}{2 \cdot 2}, x_{2} = \frac{- 7 - 71 i}{2 \cdot 2}

x = \frac{- 7 + 71 i}{2 \cdot 2} : - \frac{7}{4} + i \frac{71}{4}

x = \frac{- 7 - 71 i}{2 \cdot 2} : - \frac{7}{4} - i \frac{71}{4}

The solutions to the quadratic equation are:

x = - \frac{7}{4} + i \frac{71}{4}, x = - \frac{7}{4} - i \frac{71}{4}

\frac{1}{16} \cdot 3

7 x + 15 = 1

(- 216 x^{- 9})^{\frac{1}{3}}

- 1 \leq \frac{9 - 2 x}{5} < 4

\frac{2 π}{\frac{π}{3}}

What are the solutions to the equation 2x^2+7x+15=0 ?
The solutions to the equation 2x^2+7x+15=0 are x=-7/4+i(sqrt(71))/4 ,x=-7/4-i(sqrt(71))/4
Find the zeros of 2x^2+7x+15=0
The zeros of 2x^2+7x+15=0 are x=-7/4+i(sqrt(71))/4 ,x=-7/4-i(sqrt(71))/4