w^2+70w-40=0

Solution for w^2+70w-40=0 equation:

w^2+70w-40=0

a = 1; b = 70; c = -40;
Δ = b2-4ac
Δ = 702-4·1·(-40)
Δ = 5060
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:

$\sqrt{\Delta}=\sqrt{5060}=\sqrt{4*1265}=\sqrt{4}*\sqrt{1265}=2\sqrt{1265}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-2\sqrt{1265}}{2*1}=\frac{-70-2\sqrt{1265}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+2\sqrt{1265}}{2*1}=\frac{-70+2\sqrt{1265}}{2}
$