30=(3u-24)(u+42)

Solution for 30=(3u-24)(u+42) equation:

30=(3u-24)(u+42)

We move all terms to the left:

30-((3u-24)(u+42))=0

We multiply parentheses ..

-((+3u^2+126u-24u-1008))+30=0


We calculate terms in parentheses: -((+3u^2+126u-24u-1008)), so:

(+3u^2+126u-24u-1008)

We get rid of parentheses

3u^2+126u-24u-1008

We add all the numbers together, and all the variables

3u^2+102u-1008

Back to the equation:

-(3u^2+102u-1008)


We get rid of parentheses

-3u^2-102u+1008+30=0

We add all the numbers together, and all the variables

-3u^2-102u+1038=0

a = -3; b = -102; c = +1038;
Δ = b2-4ac
Δ = -1022-4·(-3)·1038
Δ = 22860
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:

$\sqrt{\Delta}=\sqrt{22860}=\sqrt{36*635}=\sqrt{36}*\sqrt{635}=6\sqrt{635}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-102)-6\sqrt{635}}{2*-3}=\frac{102-6\sqrt{635}}{-6}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-102)+6\sqrt{635}}{2*-3}=\frac{102+6\sqrt{635}}{-6}
$