Question

(y^2 -3y )^2 + 4 ( y^2 -3y-4)-16​

Answer 1

Answer:

y=-1,4 or y=(3-√23i)/2 , (3+√23i) /2

Explanation :

Take

${y}^{2} - 3y = t \\ then \\ {t}^{2} + 4(t - 4) - 16 = 0 \\ {t}^{2} + 4t - 32 = 0 \\ on \: solving \: this \: quadratic \: \\ equation \\ t = 4 \: or \: t = - 8 \\ put \: value \: of \: t \: nd \: \\ solve \: the \: quadrtic \: equation$