18x^2+9x-5=(3x-1)

Solution for 18x^2+9x-5=(3x-1) equation:

18x^2+9x-5=(3x-1)

We move all terms to the left:

18x^2+9x-5-((3x-1))=0


We calculate terms in parentheses: -((3x-1)), so:

(3x-1)

We get rid of parentheses

3x-1

Back to the equation:

-(3x-1)


We get rid of parentheses

18x^2+9x-3x+1-5=0

We add all the numbers together, and all the variables

18x^2+6x-4=0

a = 18; b = 6; c = -4;
Δ = b2-4ac
Δ = 62-4·18·(-4)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-18}{2*18}=\frac{-24}{36}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+18}{2*18}=\frac{12}{36}
=1/3
$