5y(y-4)=7(2y+1)

Solution for 5y(y-4)=7(2y+1) equation:

5y(y-4)=7(2y+1)

We move all terms to the left:

5y(y-4)-(7(2y+1))=0

We multiply parentheses

5y^2-20y-(7(2y+1))=0


We calculate terms in parentheses: -(7(2y+1)), so:

7(2y+1)

We multiply parentheses

14y+7

Back to the equation:

-(14y+7)


We get rid of parentheses

5y^2-20y-14y-7=0

We add all the numbers together, and all the variables

5y^2-34y-7=0

a = 5; b = -34; c = -7;
Δ = b2-4ac
Δ = -342-4·5·(-7)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-36}{2*5}=\frac{-2}{10}
=-1/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+36}{2*5}=\frac{70}{10}
=7
$