(3y-4)(2y-3)=40+(y-4)(y-13)

Solution for (3y-4)(2y-3)=40+(y-4)(y-13) equation:

(3y-4)(2y-3)=40+(y-4)(y-13)

We move all terms to the left:

(3y-4)(2y-3)-(40+(y-4)(y-13))=0

We multiply parentheses ..

(+6y^2-9y-8y+12)-(40+(y-4)(y-13))=0


We calculate terms in parentheses: -(40+(y-4)(y-13)), so:

40+(y-4)(y-13)

determiningTheFunctionDomain
(y-4)(y-13)+40

We multiply parentheses ..

(+y^2-13y-4y+52)+40

We get rid of parentheses

y^2-13y-4y+52+40

We add all the numbers together, and all the variables

y^2-17y+92

Back to the equation:

-(y^2-17y+92)


We get rid of parentheses

6y^2-y^2-9y-8y+17y+12-92=0

We add all the numbers together, and all the variables

5y^2-80=0

a = 5; b = 0; c = -80;
Δ = b2-4ac
Δ = 02-4·5·(-80)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*5}=\frac{-40}{10}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*5}=\frac{40}{10}
=4
$