What are the solutions to the equation 4x^2+77=-36 ?

The solutions to the equation 4x^2+77=-36 are x=(sqrt(113))/2 i,x=-(sqrt(113))/2 i

Find the zeros of 4x^2+77=-36

The zeros of 4x^2+77=-36 are x=(sqrt(113))/2 i,x=-(sqrt(113))/2 i

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4 x^{2} + 77 = - 36

x = \frac{113}{2} i, x = - \frac{113}{2} i

Solution steps

4 x^{2} + 77 = - 36

Move

77

to the right side

4 x^{2} = - 113

Divide both sides by

4

x^{2} = - \frac{113}{4}

For

x^{2} = f (a)

the solutions are

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

x = - \frac{113}{4}, x = - - \frac{113}{4}

- \frac{113}{4} = \frac{113}{2} i

- - \frac{113}{4} = - \frac{113}{2} i

x = \frac{113}{2} i, x = - \frac{113}{2} i

w, w^{2} + 4 w = 0

9 x - 2 - 2 x (3 x - 1) = 4 x^{2}

5 y^{2} = 7 y + 6

2 s^{2} + 15 s + 40 = 0

x, x^{2} + 10 x + 24 = 0

What are the solutions to the equation 4x^2+77=-36 ?
The solutions to the equation 4x^2+77=-36 are x=(sqrt(113))/2 i,x=-(sqrt(113))/2 i
Find the zeros of 4x^2+77=-36
The zeros of 4x^2+77=-36 are x=(sqrt(113))/2 i,x=-(sqrt(113))/2 i