0=21x^2+16x-21

Solution for 0=21x^2+16x-21 equation:

0=21x^2+16x-21

We move all terms to the left:

0-(21x^2+16x-21)=0

We add all the numbers together, and all the variables

-(21x^2+16x-21)=0

We get rid of parentheses

-21x^2-16x+21=0

a = -21; b = -16; c = +21;
Δ = b2-4ac
Δ = -162-4·(-21)·21
Δ = 2020
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:

$\sqrt{\Delta}=\sqrt{2020}=\sqrt{4*505}=\sqrt{4}*\sqrt{505}=2\sqrt{505}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{505}}{2*-21}=\frac{16-2\sqrt{505}}{-42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{505}}{2*-21}=\frac{16+2\sqrt{505}}{-42}
$