w^2+4w-130=0

Solution for w^2+4w-130=0 equation:

w^2+4w-130=0

a = 1; b = 4; c = -130;
Δ = b2-4ac
Δ = 42-4·1·(-130)
Δ = 536
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:

$\sqrt{\Delta}=\sqrt{536}=\sqrt{4*134}=\sqrt{4}*\sqrt{134}=2\sqrt{134}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{134}}{2*1}=\frac{-4-2\sqrt{134}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{134}}{2*1}=\frac{-4+2\sqrt{134}}{2}
$