15m^2-56m+48=0

Solution for 15m^2-56m+48=0 equation:

15m^2-56m+48=0

a = 15; b = -56; c = +48;
Δ = b2-4ac
Δ = -562-4·15·48
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-16}{2*15}=\frac{40}{30}
=1+1/3
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+16}{2*15}=\frac{72}{30}
=2+2/5
$