What are the solutions to the equation 15x^2+12=-8x ?

The solutions to the equation 15x^2+12=-8x are x=-4/15+i(2sqrt(41))/(15),x=-4/15-i(2sqrt(41))/(15)

Find the zeros of 15x^2+12=-8x

The zeros of 15x^2+12=-8x are x=-4/15+i(2sqrt(41))/(15),x=-4/15-i(2sqrt(41))/(15)

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15 x^{2} + 12 = - 8 x

x = - \frac{4}{15} + i \frac{2 41}{15}, x = - \frac{4}{15} - i \frac{2 41}{15}

Solution steps

15 x^{2} + 12 = - 8 x

Move

8 x

to the left side

15 x^{2} + 12 + 8 x = 0

Write in the standard form

a x^{2} + b x + c = 0

15 x^{2} + 8 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 15 \cdot 12}{2 \cdot 15}

Simplify

8^{2} - 4 \cdot 15 \cdot 12 : 441 i

x_{1, 2} = \frac{- 8 \pm 4 41 i}{2 \cdot 15}

Separate the solutions

x_{1} = \frac{- 8 + 4 41 i}{2 \cdot 15}, x_{2} = \frac{- 8 - 4 41 i}{2 \cdot 15}

x = \frac{- 8 + 4 41 i}{2 \cdot 15} : - \frac{4}{15} + i \frac{2 41}{15}

x = \frac{- 8 - 4 41 i}{2 \cdot 15} : - \frac{4}{15} - i \frac{2 41}{15}

The solutions to the quadratic equation are:

x = - \frac{4}{15} + i \frac{2 41}{15}, x = - \frac{4}{15} - i \frac{2 41}{15}

x^{2} - 3 x = 12 - 3 (x - 2)

x^{2} + 120 x + 10000 = 0

x^{2} - 2 (1 + 3 m) x + 7 (3 + 2 m) = 0

1890 = 30 x - 0.1 x^{2}

5 x^{2} + 5 = 20

What are the solutions to the equation 15x^2+12=-8x ?
The solutions to the equation 15x^2+12=-8x are x=-4/15+i(2sqrt(41))/(15),x=-4/15-i(2sqrt(41))/(15)
Find the zeros of 15x^2+12=-8x
The zeros of 15x^2+12=-8x are x=-4/15+i(2sqrt(41))/(15),x=-4/15-i(2sqrt(41))/(15)