(2*x^2)-(42*x)+160=0

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



(2x^2)-(42x)+160=0

a = 2; b = -42; c = +160;
Δ = b2-4ac
Δ = -422-4·2·160
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-22}{2*2}=\frac{20}{4}
=5
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+22}{2*2}=\frac{64}{4}
=16
$

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