21=10x^2-x

Solution for 21=10x^2-x equation:

21=10x^2-x

We move all terms to the left:

21-(10x^2-x)=0

We get rid of parentheses

-10x^2+x+21=0

a = -10; b = 1; c = +21;
Δ = b2-4ac
Δ = 12-4·(-10)·21
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-29}{2*-10}=\frac{-30}{-20}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+29}{2*-10}=\frac{28}{-20}
=-1+2/5
$