x^2+42x+424=0

Solution for x^2+42x+424=0 equation:

x^2+42x+424=0

a = 1; b = 42; c = +424;
Δ = b2-4ac
Δ = 422-4·1·424
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{17}}{2*1}=\frac{-42-2\sqrt{17}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{17}}{2*1}=\frac{-42+2\sqrt{17}}{2}
$