5(2+w)w=10+4(w+1)

Solution for 5(2+w)w=10+4(w+1) equation:

5(2+w)w=10+4(w+1)

We move all terms to the left:

5(2+w)w-(10+4(w+1))=0

We add all the numbers together, and all the variables

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

We multiply parentheses

5w^2+10w-(10+4(w+1))=0


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

10+4(w+1)

determiningTheFunctionDomain
4(w+1)+10

We multiply parentheses

4w+4+10

We add all the numbers together, and all the variables

4w+14

Back to the equation:

-(4w+14)


We get rid of parentheses

5w^2+10w-4w-14=0

We add all the numbers together, and all the variables

5w^2+6w-14=0

a = 5; b = 6; c = -14;
Δ = b2-4ac
Δ = 62-4·5·(-14)
Δ = 316
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:

$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{79}}{2*5}=\frac{-6-2\sqrt{79}}{10}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{79}}{2*5}=\frac{-6+2\sqrt{79}}{10}
$