22x^2-50x-48=0

Solution for 22x^2-50x-48=0 equation:

22x^2-50x-48=0

a = 22; b = -50; c = -48;
Δ = b2-4ac
Δ = -502-4·22·(-48)
Δ = 6724
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6724}=82$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-82}{2*22}=\frac{-32}{44}
=-8/11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+82}{2*22}=\frac{132}{44}
=3
$