10x^2-104x+150=0

Solution for 10x^2-104x+150=0 equation:

10x^2-104x+150=0

a = 10; b = -104; c = +150;
Δ = b2-4ac
Δ = -1042-4·10·150
Δ = 4816
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:

$\sqrt{\Delta}=\sqrt{4816}=\sqrt{16*301}=\sqrt{16}*\sqrt{301}=4\sqrt{301}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-104)-4\sqrt{301}}{2*10}=\frac{104-4\sqrt{301}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-104)+4\sqrt{301}}{2*10}=\frac{104+4\sqrt{301}}{20}
$