(u^2-3u+2)/(2u^2+14u+16)=0

Solution for (u^2-3u+2)/(2u^2+14u+16)=0 equation:

(u^2-3u+2)/(2u^2+14u+16)=0


Domain of the equation: (2u^2+14u+16)!=0

We move all terms containing u to the left, all other terms to the right

2u^2+14u!=-16

u∈R


We multiply all the terms by the denominator


(u^2-3u+2)=0

We get rid of parentheses

u^2-3u+2=0

a = 1; b = -3; c = +2;
Δ = b2-4ac
Δ = -32-4·1·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2*1}=\frac{2}{2}
=1
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*1}=\frac{4}{2}
=2
$