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 = 1124
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:

$\sqrt{\Delta}=\sqrt{1124}=\sqrt{4*281}=\sqrt{4}*\sqrt{281}=2\sqrt{281}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{281}}{2*1}=\frac{42-2\sqrt{281}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{281}}{2*1}=\frac{42+2\sqrt{281}}{2}
$