(3/10)y+y=36

Solution for (3/10)y+y=36 equation:

(3/10)y+y=36

We move all terms to the left:

(3/10)y+y-(36)=0


Domain of the equation: 10)y!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

(+3/10)y+y-36=0

We add all the numbers together, and all the variables

y+(+3/10)y-36=0

We multiply parentheses

3y^2+y-36=0

a = 3; b = 1; c = -36;
Δ = b2-4ac
Δ = 12-4·3·(-36)
Δ = 433
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{-(1)-\sqrt{433}}{2*3}=\frac{-1-\sqrt{433}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{433}}{2*3}=\frac{-1+\sqrt{433}}{6}
$