(2/5)w+(4/5)w-4=1

Solution for (2/5)w+(4/5)w-4=1 equation:

(2/5)w+(4/5)w-4=1

We move all terms to the left:

(2/5)w+(4/5)w-4-(1)=0


Domain of the equation: 5)w!=0

w!=0/1

w!=0

w∈R


We add all the numbers together, and all the variables

(+2/5)w+(+4/5)w-4-1=0

We add all the numbers together, and all the variables

(+2/5)w+(+4/5)w-5=0

We multiply parentheses

2w^2+4w^2-5=0

We add all the numbers together, and all the variables

6w^2-5=0

a = 6; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·6·(-5)
Δ = 120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{30}}{2*6}=\frac{0-2\sqrt{30}}{12}
=-\frac{2\sqrt{30}}{12}
=-\frac{\sqrt{30}}{6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{30}}{2*6}=\frac{0+2\sqrt{30}}{12}
=\frac{2\sqrt{30}}{12}
=\frac{\sqrt{30}}{6}
$