10=2x^2-x

Solution for 10=2x^2-x equation:

10=2x^2-x

We move all terms to the left:

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

We get rid of parentheses

-2x^2+x+10=0

a = -2; b = 1; c = +10;
Δ = b2-4ac
Δ = 12-4·(-2)·10
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*-2}=\frac{-10}{-4}
=2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*-2}=\frac{8}{-4}
=-2
$