(4y-6)(2+y)=0

Solution for (4y-6)(2+y)=0 equation:

(4y-6)(2+y)=0

We add all the numbers together, and all the variables

(4y-6)(y+2)=0

We multiply parentheses ..

(+4y^2+8y-6y-12)=0

We get rid of parentheses

4y^2+8y-6y-12=0

We add all the numbers together, and all the variables

4y^2+2y-12=0

a = 4; b = 2; c = -12;
Δ = b2-4ac
Δ = 22-4·4·(-12)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*4}=\frac{-16}{8}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*4}=\frac{12}{8}
=1+1/2
$