w^2+10w-65=0

Solution for w^2+10w-65=0 equation:

w^2+10w-65=0

a = 1; b = 10; c = -65;
Δ = b2-4ac
Δ = 102-4·1·(-65)
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{10}}{2*1}=\frac{-10-6\sqrt{10}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{10}}{2*1}=\frac{-10+6\sqrt{10}}{2}
$