2x^2-60x+420=0

Solution for 2x^2-60x+420=0 equation:

2x^2-60x+420=0

a = 2; b = -60; c = +420;
Δ = b2-4ac
Δ = -602-4·2·420
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-4\sqrt{15}}{2*2}=\frac{60-4\sqrt{15}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+4\sqrt{15}}{2*2}=\frac{60+4\sqrt{15}}{4}
$