(-w^2+3w-5)+(7w^2+6w+2)=0

Solution for (-w^2+3w-5)+(7w^2+6w+2)=0 equation:

(-w^2+3w-5)+(7w^2+6w+2)=0

We get rid of parentheses

-w^2+7w^2+3w+6w-5+2=0

We add all the numbers together, and all the variables

6w^2+9w-3=0

a = 6; b = 9; c = -3;
Δ = b2-4ac
Δ = 92-4·6·(-3)
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{17}}{2*6}=\frac{-9-3\sqrt{17}}{12}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{17}}{2*6}=\frac{-9+3\sqrt{17}}{12}
$