40=6x^2+x

Solution for 40=6x^2+x equation:

40=6x^2+x

We move all terms to the left:

40-(6x^2+x)=0

We get rid of parentheses

-6x^2-x+40=0

We add all the numbers together, and all the variables

-6x^2-1x+40=0

a = -6; b = -1; c = +40;
Δ = b2-4ac
Δ = -12-4·(-6)·40
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-31}{2*-6}=\frac{-30}{-12}
=2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+31}{2*-6}=\frac{32}{-12}
=-2+2/3
$