What are the solutions to the equation 4x^2+20=0 ?

The solutions to the equation 4x^2+20=0 are x=sqrt(5)i,x=-sqrt(5)i

Find the zeros of 4x^2+20=0

The zeros of 4x^2+20=0 are x=sqrt(5)i,x=-sqrt(5)i

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4 x^{2} + 20 = 0

x = 5 i, x = - 5 i

Solution steps

4 x^{2} + 20 = 0

Move

20

to the right side

4 x^{2} = - 20

Divide both sides by

4

x^{2} = - 5

For

x^{2} = f (a)

the solutions are

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

x = - 5, x = - - 5

- 5 = 5 i

- - 5 = - 5 i

x = 5 i, x = - 5 i

lo g_{12} (1)

- (190 \cdot 1 \cdot (- 35)) = 104 \cdot 15 \cdot 0.5 + 104 \cdot 80

\frac{5}{6} - \frac{2}{5}

x^{3} + 8 x^{2} + 19 x + 12

(4 x^{6} y^{8})^{\frac{3}{2}}

What are the solutions to the equation 4x^2+20=0 ?
The solutions to the equation 4x^2+20=0 are x=sqrt(5)i,x=-sqrt(5)i
Find the zeros of 4x^2+20=0
The zeros of 4x^2+20=0 are x=sqrt(5)i,x=-sqrt(5)i