-4/5y-3=10/7y+13

Solution for -4/5y-3=10/7y+13 equation:

-4/5y-3=10/7y+13

We move all terms to the left:

-4/5y-3-(10/7y+13)=0


Domain of the equation: 5y!=0

y!=0/5

y!=0

y∈R



Domain of the equation: 7y+13)!=0

y∈R


We get rid of parentheses

-4/5y-10/7y-13-3=0

We calculate fractions


(-28y)/35y^2+(-50y)/35y^2-13-3=0

We add all the numbers together, and all the variables

(-28y)/35y^2+(-50y)/35y^2-16=0

We multiply all the terms by the denominator


(-28y)+(-50y)-16*35y^2=0

Wy multiply elements

-560y^2+(-28y)+(-50y)=0

We get rid of parentheses

-560y^2-28y-50y=0

We add all the numbers together, and all the variables

-560y^2-78y=0

a = -560; b = -78; c = 0;
Δ = b2-4ac
Δ = -782-4·(-560)·0
Δ = 6084
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6084}=78$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-78)-78}{2*-560}=\frac{0}{-1120}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-78)+78}{2*-560}=\frac{156}{-1120}
=-39/280
$