403=w(2w+5)

Solution for 403=w(2w+5) equation:

403=w(2w+5)

We move all terms to the left:

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


We calculate terms in parentheses: -(w(2w+5)), so:

w(2w+5)

We multiply parentheses

2w^2+5w

Back to the equation:

-(2w^2+5w)


We get rid of parentheses

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

a = -2; b = -5; c = +403;
Δ = b2-4ac
Δ = -52-4·(-2)·403
Δ = 3249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3249}=57$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-57}{2*-2}=\frac{-52}{-4}
=+13
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+57}{2*-2}=\frac{62}{-4}
=-15+1/2
$