What are the solutions to the equation 2x^2+12x+50=0 ?

The solutions to the equation 2x^2+12x+50=0 are x=-3+4i,x=-3-4i

Find the zeros of 2x^2+12x+50=0

The zeros of 2x^2+12x+50=0 are x=-3+4i,x=-3-4i

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2 x^{2} + 12 x + 50 = 0

x = - 3 + 4 i, x = - 3 - 4 i

Solution steps

2 x^{2} + 12 x + 50 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 \cdot 50}{2 \cdot 2}

Simplify

1 2^{2} - 4 \cdot 2 \cdot 50 : 16 i

x_{1, 2} = \frac{- 12 \pm 16 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 12 + 16 i}{2 \cdot 2}, x_{2} = \frac{- 12 - 16 i}{2 \cdot 2}

x = \frac{- 12 + 16 i}{2 \cdot 2} : - 3 + 4 i

x = \frac{- 12 - 16 i}{2 \cdot 2} : - 3 - 4 i

The solutions to the quadratic equation are:

x = - 3 + 4 i, x = - 3 - 4 i

(x - 3)^{2} \geq 0

\frac{3}{8 ( 4 ) ^{\frac{5}{2}}}

3 x^{2} + 9 = 0

\frac{x ^{2} + 7}{x - 2} \leq \frac{- 8 x}{x - 2}

∣ x ∣ > 14

What are the solutions to the equation 2x^2+12x+50=0 ?
The solutions to the equation 2x^2+12x+50=0 are x=-3+4i,x=-3-4i
Find the zeros of 2x^2+12x+50=0
The zeros of 2x^2+12x+50=0 are x=-3+4i,x=-3-4i