21x^2-12=10x+9

Solution for 21x^2-12=10x+9 equation:

21x^2-12=10x+9

We move all terms to the left:

21x^2-12-(10x+9)=0

We get rid of parentheses

21x^2-10x-9-12=0

We add all the numbers together, and all the variables

21x^2-10x-21=0

a = 21; b = -10; c = -21;
Δ = b2-4ac
Δ = -102-4·21·(-21)
Δ = 1864
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1864}=\sqrt{4*466}=\sqrt{4}*\sqrt{466}=2\sqrt{466}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{466}}{2*21}=\frac{10-2\sqrt{466}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{466}}{2*21}=\frac{10+2\sqrt{466}}{42}
$