(-9/4)v+(4/5)=(7/8)

Solution for (-9/4)v+(4/5)=(7/8) equation:

(-9/4)v+(4/5)=(7/8)

We move all terms to the left:

(-9/4)v+(4/5)-((7/8))=0


Domain of the equation: 4)v!=0

v!=0/1

v!=0

v∈R


We add all the numbers together, and all the variables

(-9/4)v+(+4/5)-((+7/8))=0

We multiply parentheses

-9v^2+(+4/5)-((+7/8))=0

We get rid of parentheses

-9v^2+4/5-((+7/8))=0

We calculate fractions


-9v^2+()/()+()/()=0

We add all the numbers together, and all the variables

-9v^2+2=0

a = -9; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-9)·2
Δ = 72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*-9}=\frac{0-6\sqrt{2}}{-18}
=-\frac{6\sqrt{2}}{-18}
=-\frac{\sqrt{2}}{-3}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*-9}=\frac{0+6\sqrt{2}}{-18}
=\frac{6\sqrt{2}}{-18}
=\frac{\sqrt{2}}{-3}
$