0.25f(4f-3)=0.05(10f-9)

Solution for 0.25f(4f-3)=0.05(10f-9) equation:

0.25f(4f-3)=0.05(10f-9)

We move all terms to the left:

0.25f(4f-3)-(0.05(10f-9))=0

We multiply parentheses

0f^2+0f-(0.05(10f-9))=0


We calculate terms in parentheses: -(0.05(10f-9)), so:

0.05(10f-9)

We multiply parentheses

0.5f-0.45

Back to the equation:

-(0.5f-0.45)


We add all the numbers together, and all the variables

f^2+f-(0.5f-0.45)=0

We get rid of parentheses

f^2+f-0.5f+0.45=0

We add all the numbers together, and all the variables

f^2+0.5f+0.45=0

a = 1; b = 0.5; c = +0.45;
Δ = b2-4ac
Δ = 0.52-4·1·0.45
Δ = -1.55
Delta is less than zero, so there is no solution for the equation