3/2k=9k+5

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

3/2k=9k+5

We move all terms to the left:

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


Domain of the equation: 2k!=0

k!=0/2

k!=0

k∈R


We get rid of parentheses

3/2k-9k-5=0

We multiply all the terms by the denominator


-9k*2k-5*2k+3=0

Wy multiply elements

-18k^2-10k+3=0

a = -18; b = -10; c = +3;
Δ = b2-4ac
Δ = -102-4·(-18)·3
Δ = 316
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{79}}{2*-18}=\frac{10-2\sqrt{79}}{-36}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{79}}{2*-18}=\frac{10+2\sqrt{79}}{-36}
$