w^2+14w-21=0

Solution for w^2+14w-21=0 equation:

w^2+14w-21=0

a = 1; b = 14; c = -21;
Δ = b2-4ac
Δ = 142-4·1·(-21)
Δ = 280
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:

$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{70}}{2*1}=\frac{-14-2\sqrt{70}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{70}}{2*1}=\frac{-14+2\sqrt{70}}{2}
$