15m^2+16m-48=0

Solution for 15m^2+16m-48=0 equation:

15m^2+16m-48=0

a = 15; b = 16; c = -48;
Δ = b2-4ac
Δ = 162-4·15·(-48)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-56}{2*15}=\frac{-72}{30}
=-2+2/5
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+56}{2*15}=\frac{40}{30}
=1+1/3
$