/5y+7=8y-3

Solution for /5y+7=8y-3 equation:

/5y+7=8y-3

We move all terms to the left:

/5y+7-(8y-3)=0


Domain of the equation: 5y!=0

y!=0/5

y!=0

y∈R


We get rid of parentheses

/5y-8y+3+7=0

We multiply all the terms by the denominator


-8y*5y+3*5y+7*5y+=0

We add all the numbers together, and all the variables

-8y*5y+3*5y+7*5y=0

Wy multiply elements

-40y^2+15y+35y=0

We add all the numbers together, and all the variables

-40y^2+50y=0

a = -40; b = 50; c = 0;
Δ = b2-4ac
Δ = 502-4·(-40)·0
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-50}{2*-40}=\frac{-100}{-80}
=1+1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+50}{2*-40}=\frac{0}{-80}
=0
$