5w^2=27w=10

Solution for 5w^2=27w=10 equation:

5w^2=27w=10

We move all terms to the left:

5w^2-(27w)=0

a = 5; b = -27; c = 0;
Δ = b2-4ac
Δ = -272-4·5·0
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{-(-27)-27}{2*5}=\frac{0}{10}
=0
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+27}{2*5}=\frac{54}{10}
=5+2/5
$