Y=1/3x-102x+Y=4

Solution for Y=1/3x-102x+Y=4 equation:

=1/3Y-102Y+=4

We move all terms to the left:

-(1/3Y-102Y+)=0


Domain of the equation: 3Y-102Y+)!=0

Y∈R


We add all the numbers together, and all the variables

-(-102Y+1/3Y)=0

We get rid of parentheses

102Y-1/3Y=0

We multiply all the terms by the denominator


102Y*3Y-1=0

Wy multiply elements

306Y^2-1=0

a = 306; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·306·(-1)
Δ = 1224
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:

$\sqrt{\Delta}=\sqrt{1224}=\sqrt{36*34}=\sqrt{36}*\sqrt{34}=6\sqrt{34}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{34}}{2*306}=\frac{0-6\sqrt{34}}{612}
=-\frac{6\sqrt{34}}{612}
=-\frac{\sqrt{34}}{102}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{34}}{2*306}=\frac{0+6\sqrt{34}}{612}
=\frac{6\sqrt{34}}{612}
=\frac{\sqrt{34}}{102}
$