4u^2-16u+9=0

Solution for 4u^2-16u+9=0 equation:

4u^2-16u+9=0

a = 4; b = -16; c = +9;
Δ = b2-4ac
Δ = -162-4·4·9
Δ = 112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{7}}{2*4}=\frac{16-4\sqrt{7}}{8}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{7}}{2*4}=\frac{16+4\sqrt{7}}{8}
$