5w(4w+3)=12w+9

Solution for 5w(4w+3)=12w+9 equation:

5w(4w+3)=12w+9

We move all terms to the left:

5w(4w+3)-(12w+9)=0

We multiply parentheses

20w^2+15w-(12w+9)=0

We get rid of parentheses

20w^2+15w-12w-9=0

We add all the numbers together, and all the variables

20w^2+3w-9=0

a = 20; b = 3; c = -9;
Δ = b2-4ac
Δ = 32-4·20·(-9)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-27}{2*20}=\frac{-30}{40}
=-3/4
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+27}{2*20}=\frac{24}{40}
=3/5
$