5(y-3)=2y-(y+4)5y-3

Solution for 5(y-3)=2y-(y+4)5y-3 equation:

5(y-3)=2y-(y+4)5y-3

We move all terms to the left:

5(y-3)-(2y-(y+4)5y-3)=0

We multiply parentheses

5y-(2y-(y+4)5y-3)-15=0


We calculate terms in parentheses: -(2y-(y+4)5y-3), so:

2y-(y+4)5y-3

We multiply parentheses

-5y^2+2y-20y-3

We add all the numbers together, and all the variables

-5y^2-18y-3

Back to the equation:

-(-5y^2-18y-3)


We get rid of parentheses

5y^2+18y+5y+3-15=0

We add all the numbers together, and all the variables

5y^2+23y-12=0

a = 5; b = 23; c = -12;
Δ = b2-4ac
Δ = 232-4·5·(-12)
Δ = 769
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{-(23)-\sqrt{769}}{2*5}=\frac{-23-\sqrt{769}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{769}}{2*5}=\frac{-23+\sqrt{769}}{10}
$