4(u-3)(u-2)+5(u-2)=-3(u-3)

Solution for 4(u-3)(u-2)+5(u-2)=-3(u-3) equation:

4(u-3)(u-2)+5(u-2)=-3(u-3)

We move all terms to the left:

4(u-3)(u-2)+5(u-2)-(-3(u-3))=0

We multiply parentheses

4(u-3)(u-2)+5u-(-3(u-3))-10=0

We multiply parentheses ..

4(+u^2-2u-3u+6)+5u-(-3(u-3))-10=0


We calculate terms in parentheses: -(-3(u-3)), so:

-3(u-3)

We multiply parentheses

-3u+9

Back to the equation:

-(-3u+9)


We multiply parentheses

4u^2-8u-12u+5u-(-3u+9)+24-10=0

We get rid of parentheses

4u^2-8u-12u+5u+3u-9+24-10=0

We add all the numbers together, and all the variables

4u^2-12u+5=0

a = 4; b = -12; c = +5;
Δ = b2-4ac
Δ = -122-4·4·5
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8}{2*4}=\frac{4}{8}
=1/2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8}{2*4}=\frac{20}{8}
=2+1/2
$