40x^2-104x+48=0

Solution for 40x^2-104x+48=0 equation:

40x^2-104x+48=0

a = 40; b = -104; c = +48;
Δ = b2-4ac
Δ = -1042-4·40·48
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-104)-56}{2*40}=\frac{48}{80}
=3/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-104)+56}{2*40}=\frac{160}{80}
=2
$