(3/4)k+(1/2)=8

Solution for (3/4)k+(1/2)=8 equation:

(3/4)k+(1/2)=8

We move all terms to the left:

(3/4)k+(1/2)-(8)=0


Domain of the equation: 4)k!=0

k!=0/1

k!=0

k∈R


determiningTheFunctionDomain
(3/4)k-8+(1/2)=0

We add all the numbers together, and all the variables

(+3/4)k-8+(+1/2)=0

We multiply parentheses

3k^2-8+(+1/2)=0

We get rid of parentheses

3k^2-8+1/2=0

We multiply all the terms by the denominator


3k^2*2+1-8*2=0

We add all the numbers together, and all the variables

3k^2*2-15=0

Wy multiply elements

6k^2-15=0

a = 6; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·6·(-15)
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{10}}{2*6}=\frac{0-6\sqrt{10}}{12}
=-\frac{6\sqrt{10}}{12}
=-\frac{\sqrt{10}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{10}}{2*6}=\frac{0+6\sqrt{10}}{12}
=\frac{6\sqrt{10}}{12}
=\frac{\sqrt{10}}{2}
$