3x^2-x=6+16x

Solution for 3x^2-x=6+16x equation:

3x^2-x=6+16x

We move all terms to the left:

3x^2-x-(6+16x)=0

We add all the numbers together, and all the variables

3x^2-x-(16x+6)=0

We add all the numbers together, and all the variables

3x^2-1x-(16x+6)=0

We get rid of parentheses

3x^2-1x-16x-6=0

We add all the numbers together, and all the variables

3x^2-17x-6=0

a = 3; b = -17; c = -6;
Δ = b2-4ac
Δ = -172-4·3·(-6)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-19}{2*3}=\frac{-2}{6}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+19}{2*3}=\frac{36}{6}
=6
$