3(w+2)w=4(w-1)+10

Solution for 3(w+2)w=4(w-1)+10 equation:

3(w+2)w=4(w-1)+10

We move all terms to the left:

3(w+2)w-(4(w-1)+10)=0

We multiply parentheses

3w^2+6w-(4(w-1)+10)=0


We calculate terms in parentheses: -(4(w-1)+10), so:

4(w-1)+10

We multiply parentheses

4w-4+10

We add all the numbers together, and all the variables

4w+6

Back to the equation:

-(4w+6)


We get rid of parentheses

3w^2+6w-4w-6=0

We add all the numbers together, and all the variables

3w^2+2w-6=0

a = 3; b = 2; c = -6;
Δ = b2-4ac
Δ = 22-4·3·(-6)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{19}}{2*3}=\frac{-2-2\sqrt{19}}{6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{19}}{2*3}=\frac{-2+2\sqrt{19}}{6}
$