3=-3y(y-5)-2(y+0.3)

Solution for 3=-3y(y-5)-2(y+0.3) equation:

3=-3y(y-5)-2(y+0.3)

We move all terms to the left:

3-(-3y(y-5)-2(y+0.3))=0


We calculate terms in parentheses: -(-3y(y-5)-2(y+0.3)), so:

-3y(y-5)-2(y+0.3)

We multiply parentheses

-3y^2+15y-2y-0.6

We add all the numbers together, and all the variables

-3y^2+13y-0.6

Back to the equation:

-(-3y^2+13y-0.6)


We get rid of parentheses

3y^2-13y+0.6+3=0

We add all the numbers together, and all the variables

3y^2-13y+3.6=0

a = 3; b = -13; c = +3.6;
Δ = b2-4ac
Δ = -132-4·3·3.6
Δ = 125.8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{125.8}}{2*3}=\frac{13-\sqrt{125.8}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{125.8}}{2*3}=\frac{13+\sqrt{125.8}}{6}
$