(x)=0.009(3x^2-5x+x+1)

Solution for (x)=0.009(3x^2-5x+x+1) equation:

(x)=0.009(3x^2-5x+x+1)

We move all terms to the left:

(x)-(0.009(3x^2-5x+x+1))=0


We calculate terms in parentheses: -(0.009(3x^2-5x+x+1)), so:

0.009(3x^2-5x+x+1)

We multiply parentheses

0.027x^2-0.045x+0.009x+0.009

We add all the numbers together, and all the variables

0.027x^2-0.036x+0.009

Back to the equation:

-(0.027x^2-0.036x+0.009)


We get rid of parentheses

-0.027x^2+x+0.036x-0.009=0

We add all the numbers together, and all the variables

-0.027x^2+1.036x-0.009=0

a = -0.027; b = 1.036; c = -0.009;
Δ = b2-4ac
Δ = 1.0362-4·(-0.027)·(-0.009)
Δ = 1.072324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1.036)-\sqrt{1.072324}}{2*-0.027}=\frac{-1.036-\sqrt{1.072324}}{-0.054}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.036)+\sqrt{1.072324}}{2*-0.027}=\frac{-1.036+\sqrt{1.072324}}{-0.054}
$