3w^2+15=29

Solution for 3w^2+15=29 equation:

3w^2+15=29

We move all terms to the left:

3w^2+15-(29)=0

We add all the numbers together, and all the variables

3w^2-14=0

a = 3; b = 0; c = -14;
Δ = b2-4ac
Δ = 02-4·3·(-14)
Δ = 168
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{42}}{2*3}=\frac{0-2\sqrt{42}}{6}
=-\frac{2\sqrt{42}}{6}
=-\frac{\sqrt{42}}{3}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{42}}{2*3}=\frac{0+2\sqrt{42}}{6}
=\frac{2\sqrt{42}}{6}
=\frac{\sqrt{42}}{3}
$