4u^2-65u+196=0

Solution for 4u^2-65u+196=0 equation:

4u^2-65u+196=0

a = 4; b = -65; c = +196;
Δ = b2-4ac
Δ = -652-4·4·196
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-65)-33}{2*4}=\frac{32}{8}
=4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+33}{2*4}=\frac{98}{8}
=12+1/4
$