2x2-20x+48=0

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Solution for 2x2-20x+48=0 equation: 



2x^2-20x+48=0

a = 2; b = -20; c = +48;
Δ = b2-4ac
Δ = -202-4·2·48
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4}{2*2}=\frac{16}{4}
=4
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4}{2*2}=\frac{24}{4}
=6
$

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