9(y-4)=-10(y+2)+1/3

Solution for 9(y-4)=-10(y+2)+1/3 equation:

9(y-4)=-10(y+2)+1/3

We move all terms to the left:

9(y-4)-(-10(y+2)+1/3)=0

We multiply parentheses

9y-(-10(y+2)+1/3)-36=0

We multiply all the terms by the denominator


9y*3)-(-10(y+2)+1-36*3)=0

We add all the numbers together, and all the variables

9y*3)-(-10(y+2)=0

We multiply parentheses

9y*3)-(-10y-20=0

Wy multiply elements

27y^2-10y-20=0

a = 27; b = -10; c = -20;
Δ = b2-4ac
Δ = -102-4·27·(-20)
Δ = 2260
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2260}=\sqrt{4*565}=\sqrt{4}*\sqrt{565}=2\sqrt{565}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{565}}{2*27}=\frac{10-2\sqrt{565}}{54}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{565}}{2*27}=\frac{10+2\sqrt{565}}{54}
$