21x=196-x2

Solution for 21x=196-x2 equation:

21x=196-x2

We move all terms to the left:

21x-(196-x2)=0

We add all the numbers together, and all the variables

-(-1x^2+196)+21x=0

We get rid of parentheses

1x^2+21x-196=0

We add all the numbers together, and all the variables

x^2+21x-196=0

a = 1; b = 21; c = -196;
Δ = b2-4ac
Δ = 212-4·1·(-196)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-35}{2*1}=\frac{-56}{2}
=-28
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+35}{2*1}=\frac{14}{2}
=7
$