48x^2+28x-40=0

Solution for 48x^2+28x-40=0 equation:

48x^2+28x-40=0

a = 48; b = 28; c = -40;
Δ = b2-4ac
Δ = 282-4·48·(-40)
Δ = 8464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8464}=92$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-92}{2*48}=\frac{-120}{96}
=-1+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+92}{2*48}=\frac{64}{96}
=2/3
$