(4y-5)(6-y)=0

Solution for (4y-5)(6-y)=0 equation:

(4y-5)(6-y)=0

We add all the numbers together, and all the variables

(4y-5)(-1y+6)=0

We multiply parentheses ..

(-4y^2+24y+5y-30)=0

We get rid of parentheses

-4y^2+24y+5y-30=0

We add all the numbers together, and all the variables

-4y^2+29y-30=0

a = -4; b = 29; c = -30;
Δ = b2-4ac
Δ = 292-4·(-4)·(-30)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-19}{2*-4}=\frac{-48}{-8}
=+6
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+19}{2*-4}=\frac{-10}{-8}
=1+1/4
$