13y(y-6)=(4y+9)-105

Solution for 13y(y-6)=(4y+9)-105 equation:

13y(y-6)=(4y+9)-105

We move all terms to the left:

13y(y-6)-((4y+9)-105)=0

We multiply parentheses

13y^2-78y-((4y+9)-105)=0


We calculate terms in parentheses: -((4y+9)-105), so:

(4y+9)-105

We get rid of parentheses

4y+9-105

We add all the numbers together, and all the variables

4y-96

Back to the equation:

-(4y-96)


We get rid of parentheses

13y^2-78y-4y+96=0

We add all the numbers together, and all the variables

13y^2-82y+96=0

a = 13; b = -82; c = +96;
Δ = b2-4ac
Δ = -822-4·13·96
Δ = 1732
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:

$\sqrt{\Delta}=\sqrt{1732}=\sqrt{4*433}=\sqrt{4}*\sqrt{433}=2\sqrt{433}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-82)-2\sqrt{433}}{2*13}=\frac{82-2\sqrt{433}}{26}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-82)+2\sqrt{433}}{2*13}=\frac{82+2\sqrt{433}}{26}
$