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

The solutions to the equation 2x^2+5x+5=0 are x=-5/4+i(sqrt(15))/4 ,x=-5/4-i(sqrt(15))/4

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

The zeros of 2x^2+5x+5=0 are x=-5/4+i(sqrt(15))/4 ,x=-5/4-i(sqrt(15))/4

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

x = - \frac{5}{4} + i \frac{15}{4}, x = - \frac{5}{4} - i \frac{15}{4}

Solution steps

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

Solve with the quadratic formula

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

Simplify

5^{2} - 4 \cdot 2 \cdot 5 : 15 i

x_{1, 2} = \frac{- 5 \pm 15 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 15 i}{2 \cdot 2}, x_{2} = \frac{- 5 - 15 i}{2 \cdot 2}

x = \frac{- 5 + 15 i}{2 \cdot 2} : - \frac{5}{4} + i \frac{15}{4}

x = \frac{- 5 - 15 i}{2 \cdot 2} : - \frac{5}{4} - i \frac{15}{4}

The solutions to the quadratic equation are:

x = - \frac{5}{4} + i \frac{15}{4}, x = - \frac{5}{4} - i \frac{15}{4}

(2 x + 1)^{2} + 7 (2 x + 1) + 12

4 + 3 (x + 2) = 10

- 16 t^{2} + 503 t + 6 = 0

7 x^{4} y (9 x^{4} y^{3})

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What are the solutions to the equation 2x^2+5x+5=0 ?
The solutions to the equation 2x^2+5x+5=0 are x=-5/4+i(sqrt(15))/4 ,x=-5/4-i(sqrt(15))/4
Find the zeros of 2x^2+5x+5=0
The zeros of 2x^2+5x+5=0 are x=-5/4+i(sqrt(15))/4 ,x=-5/4-i(sqrt(15))/4