w^2+2w-14=0

Solution for w^2+2w-14=0 equation:

w^2+2w-14=0

a = 1; b = 2; c = -14;
Δ = b2-4ac
Δ = 22-4·1·(-14)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{15}}{2*1}=\frac{-2-2\sqrt{15}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{15}}{2*1}=\frac{-2+2\sqrt{15}}{2}
$