What are the solutions to the equation 4e^2+3e=-12 ?

The solutions to the equation 4e^2+3e=-12 are e=-3/8+i(sqrt(183))/8 ,e=-3/8-i(sqrt(183))/8

Find the zeros of 4e^2+3e=-12

The zeros of 4e^2+3e=-12 are e=-3/8+i(sqrt(183))/8 ,e=-3/8-i(sqrt(183))/8

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4 e^{2} + 3 e = - 12

e = - \frac{3}{8} + i \frac{183}{8}, e = - \frac{3}{8} - i \frac{183}{8}

Solution steps

4 e^{2} + 3 e = - 12

Move

12

to the left side

4 e^{2} + 3 e + 12 = 0

Solve with the quadratic formula

e_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 4 \cdot 12}{2 \cdot 4}

Simplify

3^{2} - 4 \cdot 4 \cdot 12 : 183 i

e_{1, 2} = \frac{- 3 \pm 183 i}{2 \cdot 4}

Separate the solutions

e_{1} = \frac{- 3 + 183 i}{2 \cdot 4}, e_{2} = \frac{- 3 - 183 i}{2 \cdot 4}

e = \frac{- 3 + 183 i}{2 \cdot 4} : - \frac{3}{8} + i \frac{183}{8}

e = \frac{- 3 - 183 i}{2 \cdot 4} : - \frac{3}{8} - i \frac{183}{8}

The solutions to the quadratic equation are:

e = - \frac{3}{8} + i \frac{183}{8}, e = - \frac{3}{8} - i \frac{183}{8}

y, (x + y)^{3} = x^{3} + y^{3}

x^{2} + 8 x + 16 = - 81

2 x^{2} + 3 x = 3

x^{2} + 8 x + 16 = - 64

(x + 3)^{2} + x^{2} = 269

What are the solutions to the equation 4e^2+3e=-12 ?
The solutions to the equation 4e^2+3e=-12 are e=-3/8+i(sqrt(183))/8 ,e=-3/8-i(sqrt(183))/8
Find the zeros of 4e^2+3e=-12
The zeros of 4e^2+3e=-12 are e=-3/8+i(sqrt(183))/8 ,e=-3/8-i(sqrt(183))/8