What are the solutions to the equation 4=-3(5x+2)^2 ?

The solutions to the equation 4=-3(5x+2)^2 are x=-2/5+i(2sqrt(3))/(15),x=-2/5-i(2sqrt(3))/(15)

Find the zeros of 4=-3(5x+2)^2

The zeros of 4=-3(5x+2)^2 are x=-2/5+i(2sqrt(3))/(15),x=-2/5-i(2sqrt(3))/(15)

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4 = - 3 (5 x + 2)^{2}

x = - \frac{2}{5} + i \frac{2 3}{15}, x = - \frac{2}{5} - i \frac{2 3}{15}

Solution steps

4 = - 3 (5 x + 2)^{2}

Switch sides

- 3 (5 x + 2)^{2} = 4

Divide both sides by

- 3

(5 x + 2)^{2} = - \frac{4}{3}

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

5 x + 2 = - \frac{4}{3} : x = - \frac{2}{5} + i \frac{2 3}{15}

Solve

5 x + 2 = - - \frac{4}{3} : x = - \frac{2}{5} - i \frac{2 3}{15}

The solutions to the quadratic equation are:

x = - \frac{2}{5} + i \frac{2 3}{15}, x = - \frac{2}{5} - i \frac{2 3}{15}

2 x^{2} + 16 x - 18 = 0

0 = 2 x^{2}

14 n^{2} - 5 = 33 n

x^{2} + 5^{2} = 8^{2}

6 x - x^{2} = x^{2} - 2 x

What are the solutions to the equation 4=-3(5x+2)^2 ?
The solutions to the equation 4=-3(5x+2)^2 are x=-2/5+i(2sqrt(3))/(15),x=-2/5-i(2sqrt(3))/(15)
Find the zeros of 4=-3(5x+2)^2
The zeros of 4=-3(5x+2)^2 are x=-2/5+i(2sqrt(3))/(15),x=-2/5-i(2sqrt(3))/(15)