(1/5)w+(7/5)w-4=6

Solution for (1/5)w+(7/5)w-4=6 equation:

(1/5)w+(7/5)w-4=6

We move all terms to the left:

(1/5)w+(7/5)w-4-(6)=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

(+1/5)w+(+7/5)w-4-6=0

We add all the numbers together, and all the variables

(+1/5)w+(+7/5)w-10=0

We multiply parentheses

w^2+7w^2-10=0

We add all the numbers together, and all the variables

8w^2-10=0

a = 8; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·8·(-10)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*8}=\frac{0-8\sqrt{5}}{16}
=-\frac{8\sqrt{5}}{16}
=-\frac{\sqrt{5}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*8}=\frac{0+8\sqrt{5}}{16}
=\frac{8\sqrt{5}}{16}
=\frac{\sqrt{5}}{2}
$