factor 7x^2+35=0?

7x² + 35 = 0
7(x² + 5) = 0
(x² + 5) = 0
x² = – 5
x = ± √(–5)
± i √(5)
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7x² + 35 = 0
7x² = - 35
x² = - 5
x = - (√5)

Answer: x = (- 1)√5

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7(x^2+5)=0

x^2=-5
x=sqrt +/-5i

7x² = - 35
x² = - 5
x² = 5 i²
x = ±√5 i

7x^2 + 35 = 0
(7x^2 + 35)/7 = 0/7
x^2 + 5 = 0
x^2 = -5
x = √-5 (imaginary number)

x^2 = -35/7
x^2=-5
the roots are imaginary

x=+sqrt(5)* i, -sqrt(5)* i

7x^2+35=0
Divide by 7

x^2+5=0

x^2=-5

x= +- sqrt 5i