(2/6)k+3=10

Solution for (2/6)k+3=10 equation:

(2/6)k+3=10

We move all terms to the left:

(2/6)k+3-(10)=0


Domain of the equation: 6)k!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

(+2/6)k+3-10=0

We add all the numbers together, and all the variables

(+2/6)k-7=0

We multiply parentheses

2k^2-7=0

a = 2; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·2·(-7)
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{14}}{2*2}=\frac{0-2\sqrt{14}}{4}
=-\frac{2\sqrt{14}}{4}
=-\frac{\sqrt{14}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{14}}{2*2}=\frac{0+2\sqrt{14}}{4}
=\frac{2\sqrt{14}}{4}
=\frac{\sqrt{14}}{2}
$