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Solve : `9x^(2) + 10x + 3 = 0`.

Answer» The given equation is `9x^(2)+10x+3=0`.
This is of the form `ax^(2)+bx+c=0`, Where a = 9, b = 10 and c = 3.
`therefore" "(b^(2)-4ac)={(10)^(2)-4xx9xx3}=(100-108)=-8 lt 0`.
So, the given equation has complex roots.
These roots are given by
`(-b+- sqrt(b^(2)-4ac))/(2a)=(-10 +- sqrt(-8))/(2xx9)" "[because (b^(2)-4ac)=-8]`
`=(-10 +- i2 sqrt(2))/(18)=(-5+- i sqrt(2))/(9)`.
`therefore" ""solution set"={(-5+ i sqrt(2))/(9),(-5- i sqrt(2))/(9)}={(-5)/(9)+(sqrt(2))/(9)i, (-5)/(9)-(sqrt(2))/(9)i}`.


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