x^2-50x+150=0

Solution for x^2-50x+150=0 equation:

x^2-50x+150=0

a = 1; b = -50; c = +150;
Δ = b2-4ac
Δ = -502-4·1·150
Δ = 1900
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:

$\sqrt{\Delta}=\sqrt{1900}=\sqrt{100*19}=\sqrt{100}*\sqrt{19}=10\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-10\sqrt{19}}{2*1}=\frac{50-10\sqrt{19}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+10\sqrt{19}}{2*1}=\frac{50+10\sqrt{19}}{2}
$