U^2-2u+6=3u+2u^2

Solution for U^2-2u+6=3u+2u^2 equation:

^2-2U+6=3U+2U^2

We move all terms to the left:

^2-2U+6-(3U+2U^2)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

-2U^2-3U-2U=0

We add all the numbers together, and all the variables

-2U^2-5U=0

a = -2; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-2)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$U_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$U_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$U_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-2}=\frac{0}{-4}
=0
$
$U_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-2}=\frac{10}{-4}
=-2+1/2
$