(7/2)w-5=-6w+3

Solution for (7/2)w-5=-6w+3 equation:

(7/2)w-5=-6w+3

We move all terms to the left:

(7/2)w-5-(-6w+3)=0


Domain of the equation: 2)w!=0

w!=0/1

w!=0

w∈R


We add all the numbers together, and all the variables

(+7/2)w-(-6w+3)-5=0

We multiply parentheses

7w^2-(-6w+3)-5=0

We get rid of parentheses

7w^2+6w-3-5=0

We add all the numbers together, and all the variables

7w^2+6w-8=0

a = 7; b = 6; c = -8;
Δ = b2-4ac
Δ = 62-4·7·(-8)
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{65}}{2*7}=\frac{-6-2\sqrt{65}}{14}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{65}}{2*7}=\frac{-6+2\sqrt{65}}{14}
$