What are the solutions to the equation (4x+7)^2+81=0 ?

The solutions to the equation (4x+7)^2+81=0 are x=-7/4+i 9/4 ,x=-7/4-i 9/4

Find the zeros of (4x+7)^2+81=0

The zeros of (4x+7)^2+81=0 are x=-7/4+i 9/4 ,x=-7/4-i 9/4

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(4 x + 7)^{2} + 81 = 0

x = - \frac{7}{4} + i \frac{9}{4}, x = - \frac{7}{4} - i \frac{9}{4}

Solution steps

(4 x + 7)^{2} + 81 = 0

Move

81

to the right side

(4 x + 7)^{2} = - 81

For

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

the solutions are

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

Solve

4 x + 7 = - 81 : x = - \frac{7}{4} + i \frac{9}{4}

Solve

4 x + 7 = - - 81 : x = - \frac{7}{4} - i \frac{9}{4}

The solutions to the quadratic equation are:

x = - \frac{7}{4} + i \frac{9}{4}, x = - \frac{7}{4} - i \frac{9}{4}

\frac{8}{7 + i}

96 = k (4^{2})

\frac{n}{3} - 10 \leq - 10

4 + 2 j < - 4

\frac{x}{2} + 7 < 2

What are the solutions to the equation (4x+7)^2+81=0 ?
The solutions to the equation (4x+7)^2+81=0 are x=-7/4+i 9/4 ,x=-7/4-i 9/4
Find the zeros of (4x+7)^2+81=0
The zeros of (4x+7)^2+81=0 are x=-7/4+i 9/4 ,x=-7/4-i 9/4