x^2+42x-160=0

Solution for x^2+42x-160=0 equation:

x^2+42x-160=0

a = 1; b = 42; c = -160;
Δ = b2-4ac
Δ = 422-4·1·(-160)
Δ = 2404
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:

$\sqrt{\Delta}=\sqrt{2404}=\sqrt{4*601}=\sqrt{4}*\sqrt{601}=2\sqrt{601}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{601}}{2*1}=\frac{-42-2\sqrt{601}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{601}}{2*1}=\frac{-42+2\sqrt{601}}{2}
$