3k2+4=130

Solution for 3k2+4=130 equation:

3k^2+4=130

We move all terms to the left:

3k^2+4-(130)=0

We add all the numbers together, and all the variables

3k^2-126=0

a = 3; b = 0; c = -126;
Δ = b2-4ac
Δ = 02-4·3·(-126)
Δ = 1512
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:

$\sqrt{\Delta}=\sqrt{1512}=\sqrt{36*42}=\sqrt{36}*\sqrt{42}=6\sqrt{42}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{42}}{2*3}=\frac{0-6\sqrt{42}}{6}
=-\frac{6\sqrt{42}}{6}
=-\sqrt{42}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{42}}{2*3}=\frac{0+6\sqrt{42}}{6}
=\frac{6\sqrt{42}}{6}
=\sqrt{42}
$