29-3+10y(y+5)=5(5y-4)-6(y-1)-20y+10

Solution for 29-3+10y(y+5)=5(5y-4)-6(y-1)-20y+10 equation:

29-3+10y(y+5)=5(5y-4)-6(y-1)-20y+10

We move all terms to the left:

29-3+10y(y+5)-(5(5y-4)-6(y-1)-20y+10)=0

We add all the numbers together, and all the variables

10y(y+5)-(5(5y-4)-6(y-1)-20y+10)+26=0

We multiply parentheses

10y^2+50y-(5(5y-4)-6(y-1)-20y+10)+26=0


We calculate terms in parentheses: -(5(5y-4)-6(y-1)-20y+10), so:

5(5y-4)-6(y-1)-20y+10

We add all the numbers together, and all the variables

-20y+5(5y-4)-6(y-1)+10

We multiply parentheses

-20y+25y-6y-20+6+10

We add all the numbers together, and all the variables

-1y-4

Back to the equation:

-(-1y-4)


We get rid of parentheses

10y^2+50y+1y+4+26=0

We add all the numbers together, and all the variables

10y^2+51y+30=0

a = 10; b = 51; c = +30;
Δ = b2-4ac
Δ = 512-4·10·30
Δ = 1401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-\sqrt{1401}}{2*10}=\frac{-51-\sqrt{1401}}{20}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+\sqrt{1401}}{2*10}=\frac{-51+\sqrt{1401}}{20}
$