u^2+5u+23/4=0

Solution for u^2+5u+23/4=0 equation:

u^2+5u+23/4=0

We multiply all the terms by the denominator


u^2*4+5u*4+23=0

Wy multiply elements

4u^2+20u+23=0

a = 4; b = 20; c = +23;
Δ = b2-4ac
Δ = 202-4·4·23
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{2}}{2*4}=\frac{-20-4\sqrt{2}}{8}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{2}}{2*4}=\frac{-20+4\sqrt{2}}{8}
$