(48x^2)+140x+100=0

Solution for (48x^2)+140x+100=0 equation:

(48x^2)+140x+100=0

a = 48; b = 140; c = +100;
Δ = b2-4ac
Δ = 1402-4·48·100
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(140)-20}{2*48}=\frac{-160}{96}
=-1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(140)+20}{2*48}=\frac{-120}{96}
=-1+1/4
$