x^2-10x+21=140

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

x^2-10x+21=140

We move all terms to the left:

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

We add all the numbers together, and all the variables

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