1300=(w+3)2w

Solution for 1300=(w+3)2w equation:

1300=(w+3)2w

We move all terms to the left:

1300-((w+3)2w)=0


We calculate terms in parentheses: -((w+3)2w), so:

(w+3)2w

We multiply parentheses

2w^2+6w

Back to the equation:

-(2w^2+6w)


We get rid of parentheses

-2w^2-6w+1300=0

a = -2; b = -6; c = +1300;
Δ = b2-4ac
Δ = -62-4·(-2)·1300
Δ = 10436
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:

$\sqrt{\Delta}=\sqrt{10436}=\sqrt{4*2609}=\sqrt{4}*\sqrt{2609}=2\sqrt{2609}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{2609}}{2*-2}=\frac{6-2\sqrt{2609}}{-4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{2609}}{2*-2}=\frac{6+2\sqrt{2609}}{-4}
$