1/2y-3=1y+10

Solution for 1/2y-3=1y+10 equation:

1/2y-3=1y+10

We move all terms to the left:

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


Domain of the equation: 2y!=0

y!=0/2

y!=0

y∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

1/2y-y-10-3=0

We multiply all the terms by the denominator


-y*2y-10*2y-3*2y+1=0

Wy multiply elements

-2y^2-20y-6y+1=0

We add all the numbers together, and all the variables

-2y^2-26y+1=0

a = -2; b = -26; c = +1;
Δ = b2-4ac
Δ = -262-4·(-2)·1
Δ = 684
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:

$\sqrt{\Delta}=\sqrt{684}=\sqrt{36*19}=\sqrt{36}*\sqrt{19}=6\sqrt{19}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-6\sqrt{19}}{2*-2}=\frac{26-6\sqrt{19}}{-4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+6\sqrt{19}}{2*-2}=\frac{26+6\sqrt{19}}{-4}
$