(5w^2+9w+4)+(-2w^2+4w+4)-(5w^2+7w-5)=0

Solution for (5w^2+9w+4)+(-2w^2+4w+4)-(5w^2+7w-5)=0 equation:

(5w^2+9w+4)+(-2w^2+4w+4)-(5w^2+7w-5)=0

We get rid of parentheses

-2w^2+5w^2-5w^2+4w+9w-7w+4+4+5=0

We add all the numbers together, and all the variables

-2w^2+6w+13=0

a = -2; b = 6; c = +13;
Δ = b2-4ac
Δ = 62-4·(-2)·13
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{35}}{2*-2}=\frac{-6-2\sqrt{35}}{-4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{35}}{2*-2}=\frac{-6+2\sqrt{35}}{-4}
$