2x^2-152x+2040=0

Solution for 2x^2-152x+2040=0 equation:

2x^2-152x+2040=0

a = 2; b = -152; c = +2040;
Δ = b2-4ac
Δ = -1522-4·2·2040
Δ = 6784
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:

$\sqrt{\Delta}=\sqrt{6784}=\sqrt{64*106}=\sqrt{64}*\sqrt{106}=8\sqrt{106}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-152)-8\sqrt{106}}{2*2}=\frac{152-8\sqrt{106}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-152)+8\sqrt{106}}{2*2}=\frac{152+8\sqrt{106}}{4}
$