Question

find the zeros
y²+3/2√5y-5

Answer 1

y²+3/2√5y-5=0

Splitting the middle term
y²+4/2√5y-√5/2y-5=0
Taking LCM
2y²+4√5y-√5y-10=0       (Denominator taken to RHS)
2y(y+2√5)-√5(y+2√5)=0
(2y-√5)(y+2√5)=0
∴y= √5/2 or -2√5

Hope you understood :) 

Answer 2

We must recall that:

In algebra, a quadratic equation is any equation that can be rearranged in standard form as $ax^2+bx+c=0$ where $x$ represents an unknown, and a, b, and c represent known numbers, where $a\neq 0.$ If $a=0,$ then the equation is linear, not quadratic, as there is no $ax^2$ term.

Given: $y^2+\frac{3}{2}\sqrt{5}y-5=0$

$y^2+\frac{4}{2}\sqrt{5}y-\frac{1}{2}\sqrt{5}y-5=0\\ 2y^2+4\sqrt{5}y-\sqrt{5}y-10=0\\ 2y(y+2\sqrt{5})-\sqrt{5}(y+2\sqrt{5})=0\\ (2y-\sqrt{5})(y+2\sqrt{5})=0\\ 2y=\sqrt{5} \\ y=\frac{\sqrt{5} }{2} \\y+2\sqrt{5}=0\\ y=-2\sqrt{5}$