7w+27=5w(w+3)

Solution for 7w+27=5w(w+3) equation:

7w+27=5w(w+3)

We move all terms to the left:

7w+27-(5w(w+3))=0


We calculate terms in parentheses: -(5w(w+3)), so:

5w(w+3)

We multiply parentheses

5w^2+15w

Back to the equation:

-(5w^2+15w)


We get rid of parentheses

-5w^2+7w-15w+27=0

We add all the numbers together, and all the variables

-5w^2-8w+27=0

a = -5; b = -8; c = +27;
Δ = b2-4ac
Δ = -82-4·(-5)·27
Δ = 604
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:

$\sqrt{\Delta}=\sqrt{604}=\sqrt{4*151}=\sqrt{4}*\sqrt{151}=2\sqrt{151}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{151}}{2*-5}=\frac{8-2\sqrt{151}}{-10}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{151}}{2*-5}=\frac{8+2\sqrt{151}}{-10}
$