w^2+5w=403

Solution for w^2+5w=403 equation:

w^2+5w=403

We move all terms to the left:

w^2+5w-(403)=0

a = 1; b = 5; c = -403;
Δ = b2-4ac
Δ = 52-4·1·(-403)
Δ = 1637
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{1637}}{2*1}=\frac{-5-\sqrt{1637}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{1637}}{2*1}=\frac{-5+\sqrt{1637}}{2}
$