k(2)+k(2)+9=3(k+k)(2)-1

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

k(2)+k(2)+9=3(k+k)(2)-1

We move all terms to the left:

k(2)+k(2)+9-(3(k+k)(2)-1)=0

We add all the numbers together, and all the variables

k2+k2-(3(+2k)2-1)+9=0

We add all the numbers together, and all the variables

2k^2-(3(+2k)2-1)+9=0


We calculate terms in parentheses: -(3(+2k)2-1), so:

3(+2k)2-1

We multiply parentheses

12k-1

Back to the equation:

-(12k-1)


We get rid of parentheses

2k^2-12k+1+9=0

We add all the numbers together, and all the variables

2k^2-12k+10=0

a = 2; b = -12; c = +10;
Δ = b2-4ac
Δ = -122-4·2·10
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8}{2*2}=\frac{4}{4}
=1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8}{2*2}=\frac{20}{4}
=5
$