9/2k+-7/2=-3-4/5k

Solution for 9/2k+-7/2=-3-4/5k equation:

9/2k+-7/2=-3-4/5k

We move all terms to the left:

9/2k+-7/2-(-3-4/5k)=0


Domain of the equation: 2k!=0

k!=0/2

k!=0

k∈R



Domain of the equation: 5k)!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

9/2k-(-4/5k-3)+-7/2=0

We add all the numbers together, and all the variables

9/2k-(-4/5k-3)-7/2=0

We get rid of parentheses

9/2k+4/5k+3-7/2=0

We calculate fractions


45k/40k^2+32k/40k^2+(-35k)/40k^2+3=0

We multiply all the terms by the denominator


45k+32k+(-35k)+3*40k^2=0

We add all the numbers together, and all the variables

77k+(-35k)+3*40k^2=0

Wy multiply elements

120k^2+77k+(-35k)=0

We get rid of parentheses

120k^2+77k-35k=0

We add all the numbers together, and all the variables

120k^2+42k=0

a = 120; b = 42; c = 0;
Δ = b2-4ac
Δ = 422-4·120·0
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42}{2*120}=\frac{-84}{240}
=-7/20
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42}{2*120}=\frac{0}{240}
=0
$