4u^2+20u+9=0

Solution for 4u^2+20u+9=0 equation:

4u^2+20u+9=0

a = 4; b = 20; c = +9;
Δ = b2-4ac
Δ = 202-4·4·9
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*4}=\frac{-36}{8}
=-4+1/2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*4}=\frac{-4}{8}
=-1/2
$