2,4x^2-60x+300=0

Solution for 2,4x^2-60x+300=0 equation:

2.4x^2-60x+300=0

a = 2.4; b = -60; c = +300;
Δ = b2-4ac
Δ = -602-4·2.4·300
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-12\sqrt{5}}{2*2.4}=\frac{60-12\sqrt{5}}{4.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+12\sqrt{5}}{2*2.4}=\frac{60+12\sqrt{5}}{4.8}
$