x^2+210x+2065=0

Solution for x^2+210x+2065=0 equation:

x^2+210x+2065=0

a = 1; b = 210; c = +2065;
Δ = b2-4ac
Δ = 2102-4·1·2065
Δ = 35840
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:

$\sqrt{\Delta}=\sqrt{35840}=\sqrt{1024*35}=\sqrt{1024}*\sqrt{35}=32\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(210)-32\sqrt{35}}{2*1}=\frac{-210-32\sqrt{35}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(210)+32\sqrt{35}}{2*1}=\frac{-210+32\sqrt{35}}{2}
$