2x^2-64x+480=0

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Solution for 2x^2-64x+480=0 equation:



2x^2-64x+480=0
a = 2; b = -64; c = +480;
Δ = b2-4ac
Δ = -642-4·2·480
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-16}{2*2}=\frac{48}{4} =12 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+16}{2*2}=\frac{80}{4} =20 $

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