21n2-10=n

Solution for 21n2-10=n equation:

21n^2-10=n

We move all terms to the left:

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

We add all the numbers together, and all the variables

21n^2-1n-10=0

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

$\sqrt{\Delta}=\sqrt{841}=29$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-29}{2*21}=\frac{-28}{42}
=-2/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+29}{2*21}=\frac{30}{42}
=5/7
$