(4y-3)(6+y)=0

Solution for (4y-3)(6+y)=0 equation:

(4y-3)(6+y)=0

We add all the numbers together, and all the variables

(4y-3)(y+6)=0

We multiply parentheses ..

(+4y^2+24y-3y-18)=0

We get rid of parentheses

4y^2+24y-3y-18=0

We add all the numbers together, and all the variables

4y^2+21y-18=0

a = 4; b = 21; c = -18;
Δ = b2-4ac
Δ = 212-4·4·(-18)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-27}{2*4}=\frac{-48}{8}
=-6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+27}{2*4}=\frac{6}{8}
=3/4
$