2-3/4(k)=1/8(k)+9

Solution for 2-3/4(k)=1/8(k)+9 equation:

2-3/4(k)=1/8(k)+9

We move all terms to the left:

2-3/4(k)-(1/8(k)+9)=0


Domain of the equation: 4k!=0

k!=0/4

k!=0

k∈R



Domain of the equation: 8k+9)!=0

k∈R


We get rid of parentheses

-3/4k-1/8k-9+2=0

We calculate fractions


(-24k)/32k^2+(-4k)/32k^2-9+2=0

We add all the numbers together, and all the variables

(-24k)/32k^2+(-4k)/32k^2-7=0

We multiply all the terms by the denominator


(-24k)+(-4k)-7*32k^2=0

Wy multiply elements

-224k^2+(-24k)+(-4k)=0

We get rid of parentheses

-224k^2-24k-4k=0

We add all the numbers together, and all the variables

-224k^2-28k=0

a = -224; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·(-224)·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*-224}=\frac{0}{-448}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*-224}=\frac{56}{-448}
=-1/8
$