64u^2-232u+198=0

Solution for 64u^2-232u+198=0 equation:

64u^2-232u+198=0

a = 64; b = -232; c = +198;
Δ = b2-4ac
Δ = -2322-4·64·198
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-232)-56}{2*64}=\frac{176}{128}
=1+3/8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-232)+56}{2*64}=\frac{288}{128}
=2+1/4
$