140=x^2-10x+21

Solution for 140=x^2-10x+21 equation:

140=x^2-10x+21

We move all terms to the left:

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

We get rid of parentheses

-x^2+10x-21+140=0

We add all the numbers together, and all the variables

-1x^2+10x+119=0

a = -1; b = 10; c = +119;
Δ = b2-4ac
Δ = 102-4·(-1)·119
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-24}{2*-1}=\frac{-34}{-2}
=+17
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+24}{2*-1}=\frac{14}{-2}
=-7
$