2x^2-42x+144=0

Solution for 2x^2-42x+144=0 equation:

2x^2-42x+144=0

a = 2; b = -42; c = +144;
Δ = b2-4ac
Δ = -422-4·2·144
Δ = 612
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:

$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6\sqrt{17}}{2*2}=\frac{42-6\sqrt{17}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6\sqrt{17}}{2*2}=\frac{42+6\sqrt{17}}{4}
$