x^2+105x+50=0

Solution for x^2+105x+50=0 equation:

x^2+105x+50=0

a = 1; b = 105; c = +50;
Δ = b2-4ac
Δ = 1052-4·1·50
Δ = 10825
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:

$\sqrt{\Delta}=\sqrt{10825}=\sqrt{25*433}=\sqrt{25}*\sqrt{433}=5\sqrt{433}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(105)-5\sqrt{433}}{2*1}=\frac{-105-5\sqrt{433}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(105)+5\sqrt{433}}{2*1}=\frac{-105+5\sqrt{433}}{2}
$