2(3w+2w^2)-6=2w(w-4)+10

Solution for 2(3w+2w^2)-6=2w(w-4)+10 equation:

2(3w+2w^2)-6=2w(w-4)+10

We move all terms to the left:

2(3w+2w^2)-6-(2w(w-4)+10)=0

We multiply parentheses

4w^2+6w-(2w(w-4)+10)-6=0


We calculate terms in parentheses: -(2w(w-4)+10), so:

2w(w-4)+10

We multiply parentheses

2w^2-8w+10

Back to the equation:

-(2w^2-8w+10)


We get rid of parentheses

4w^2-2w^2+6w+8w-10-6=0

We add all the numbers together, and all the variables

2w^2+14w-16=0

a = 2; b = 14; c = -16;
Δ = b2-4ac
Δ = 142-4·2·(-16)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-18}{2*2}=\frac{-32}{4}
=-8
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+18}{2*2}=\frac{4}{4}
=1
$