x^2+21x=140

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

x^2+21x=140

We move all terms to the left:

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

a = 1; b = 21; c = -140;
Δ = b2-4ac
Δ = 212-4·1·(-140)
Δ = 1001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{1001}}{2*1}=\frac{-21-\sqrt{1001}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{1001}}{2*1}=\frac{-21+\sqrt{1001}}{2}
$