2x^2+x-3*3x-1=18x^2+3x-6

Solution for 2x^2+x-3*3x-1=18x^2+3x-6 equation:

2x^2+x-3*3x-1=18x^2+3x-6

We move all terms to the left:

2x^2+x-3*3x-1-(18x^2+3x-6)=0

Wy multiply elements

2x^2+x-9x-(18x^2+3x-6)-1=0

We get rid of parentheses

2x^2-18x^2+x-9x-3x+6-1=0

We add all the numbers together, and all the variables

-16x^2-11x+5=0

a = -16; b = -11; c = +5;
Δ = b2-4ac
Δ = -112-4·(-16)·5
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-21}{2*-16}=\frac{-10}{-32}
=5/16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+21}{2*-16}=\frac{32}{-32}
=-1
$