x^2+10x-405=0

Solution for x^2+10x-405=0 equation:

x^2+10x-405=0

a = 1; b = 10; c = -405;
Δ = b2-4ac
Δ = 102-4·1·(-405)
Δ = 1720
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:

$\sqrt{\Delta}=\sqrt{1720}=\sqrt{4*430}=\sqrt{4}*\sqrt{430}=2\sqrt{430}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{430}}{2*1}=\frac{-10-2\sqrt{430}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{430}}{2*1}=\frac{-10+2\sqrt{430}}{2}
$