3y(4y+13)=8(7y+5)

Solution for 3y(4y+13)=8(7y+5) equation:

3y(4y+13)=8(7y+5)

We move all terms to the left:

3y(4y+13)-(8(7y+5))=0

We multiply parentheses

12y^2+39y-(8(7y+5))=0


We calculate terms in parentheses: -(8(7y+5)), so:

8(7y+5)

We multiply parentheses

56y+40

Back to the equation:

-(56y+40)


We get rid of parentheses

12y^2+39y-56y-40=0

We add all the numbers together, and all the variables

12y^2-17y-40=0

a = 12; b = -17; c = -40;
Δ = b2-4ac
Δ = -172-4·12·(-40)
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-47}{2*12}=\frac{-30}{24}
=-1+1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+47}{2*12}=\frac{64}{24}
=2+2/3
$