Question

X(2x+5)=3
solve
question of quadratic.

Answer 1

$\huge\mathbb{SOLUTION:-}$

$\sf\bullet x(2x+5)=3\\ \\ \sf\implies 2x{}^{2}+5x=3\\ \\ \sf\implies 2x{}^{2}+5x-3=0\\ \\ \sf\implies 2x{}^{2}+(6-1)x-3=0\\ \\ \sf\implies 2x{}^{2}+6x-x-3=0\\ \\ \sf\implies 2x(x+3)-1(x+3)=0\\ \\ \sf\implies (x+3)(2x-1)=0$

Now the zero's of polynomial

$\sf\bullet x+3=0\\ \\ \sf\implies x=(-3)\\ \\ \sf\bullet 2x-1=0\\ \\ \sf\implies 2x-1\\ \\ \sf\implies x=\dfrac{1}{2}$

$\boxed{\sf{Zero's\:of\: Polynomial= (-3)\:\:and\:\:\dfrac{1}{2}}}$