5(y-1)=3(2y-5)(1-3y)

Solution for 5(y-1)=3(2y-5)(1-3y) equation:

5(y-1)=3(2y-5)(1-3y)

We move all terms to the left:

5(y-1)-(3(2y-5)(1-3y))=0

We add all the numbers together, and all the variables

5(y-1)-(3(2y-5)(-3y+1))=0

We multiply parentheses

5y-(3(2y-5)(-3y+1))-5=0

We multiply parentheses ..

-(3(-6y^2+2y+15y-5))+5y-5=0


We calculate terms in parentheses: -(3(-6y^2+2y+15y-5)), so:

3(-6y^2+2y+15y-5)

We multiply parentheses

-18y^2+6y+45y-15

We add all the numbers together, and all the variables

-18y^2+51y-15

Back to the equation:

-(-18y^2+51y-15)


We get rid of parentheses

18y^2-51y+5y+15-5=0

We add all the numbers together, and all the variables

18y^2-46y+10=0

a = 18; b = -46; c = +10;
Δ = b2-4ac
Δ = -462-4·18·10
Δ = 1396
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1396}=\sqrt{4*349}=\sqrt{4}*\sqrt{349}=2\sqrt{349}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-46)-2\sqrt{349}}{2*18}=\frac{46-2\sqrt{349}}{36}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-46)+2\sqrt{349}}{2*18}=\frac{46+2\sqrt{349}}{36}
$