Question
( x+a+=-frac{3}{5} )
( x+n=frac{4}{-3} )
( 9 x^{2}+27 x+20=0 )
3) ( 6 x^{2}+2 cdot 3 x cdot frac{27}{8}^{9} d x+frac{81}{4}+20-frac{81}{4}=0 )
( 1left(3 n+frac{9}{2}right)^{2}= )
2) ( left(3 n+frac{9}{2}right)^{2}=frac{1}{4} )

0.88 If an angle A of a AABC satisfies 5 cos A + 3 = 0, then the roots of the quadratic equation, 9x2 + 27x + 20 = 0 are - (JEE-Main Online-2018] (A) sin A, sec A (B) sec A, tan A (C) tan A, cos A (D) sec A, cot A
Solution
