w^2+3w=168

Solution for w^2+3w=168 equation:

w^2+3w=168

We move all terms to the left:

w^2+3w-(168)=0

a = 1; b = 3; c = -168;
Δ = b2-4ac
Δ = 32-4·1·(-168)
Δ = 681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{681}}{2*1}=\frac{-3-\sqrt{681}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{681}}{2*1}=\frac{-3+\sqrt{681}}{2}
$