10=3x2+x

Solution for 10=3x2+x equation:

10=3x^2+x

We move all terms to the left:

10-(3x^2+x)=0

We get rid of parentheses

-3x^2-x+10=0

We add all the numbers together, and all the variables

-3x^2-1x+10=0

a = -3; b = -1; c = +10;
Δ = b2-4ac
Δ = -12-4·(-3)·10
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*-3}=\frac{-10}{-6}
=1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*-3}=\frac{12}{-6}
=-2
$