algebra factoring

Question

solve by factoring 5x^2 + 10x + 20

Answer 1

Given equation 5x^2+10x+20

Compare it qudratic equation ax^2+bx+c = 0

a = 5, b = 10, c = 20

Roots are x = [-b±√(b^2-4ac)]/2a

Given system has imaginary roots.

x = [-10±√(100-4*5*20)]/2*5

x = [-10±√(100-400)]/10

x = [-10±√-300]/10

x = [-10±10√-3]/10

x = 10[1±√-3]/10

x = 1±1.73i

x = 1+1.73i      and x = 1-1.73i

Solution is x = 1+1.73i and 1-1.73i.