7/5y=7y+40

Solution for 7/5y=7y+40 equation:

7/5y=7y+40

We move all terms to the left:

7/5y-(7y+40)=0


Domain of the equation: 5y!=0

y!=0/5

y!=0

y∈R


We get rid of parentheses

7/5y-7y-40=0

We multiply all the terms by the denominator


-7y*5y-40*5y+7=0

Wy multiply elements

-35y^2-200y+7=0

a = -35; b = -200; c = +7;
Δ = b2-4ac
Δ = -2002-4·(-35)·7
Δ = 40980
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:

$\sqrt{\Delta}=\sqrt{40980}=\sqrt{4*10245}=\sqrt{4}*\sqrt{10245}=2\sqrt{10245}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-200)-2\sqrt{10245}}{2*-35}=\frac{200-2\sqrt{10245}}{-70}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-200)+2\sqrt{10245}}{2*-35}=\frac{200+2\sqrt{10245}}{-70}
$