x2+40x+21=0

Solution for x2+40x+21=0 equation:

x2+40x+21=0

We add all the numbers together, and all the variables

x^2+40x+21=0

a = 1; b = 40; c = +21;
Δ = b2-4ac
Δ = 402-4·1·21
Δ = 1516
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:

$\sqrt{\Delta}=\sqrt{1516}=\sqrt{4*379}=\sqrt{4}*\sqrt{379}=2\sqrt{379}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{379}}{2*1}=\frac{-40-2\sqrt{379}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{379}}{2*1}=\frac{-40+2\sqrt{379}}{2}
$