u^2-4u-20=3u+24=

Solution for u^2-4u-20=3u+24= equation:

u^2-4u-20=3u+24=

We move all terms to the left:

u^2-4u-20-(3u+24)=0

We get rid of parentheses

u^2-4u-3u-24-20=0

We add all the numbers together, and all the variables

u^2-7u-44=0

a = 1; b = -7; c = -44;
Δ = b2-4ac
Δ = -72-4·1·(-44)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-15}{2*1}=\frac{-8}{2}
=-4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+15}{2*1}=\frac{22}{2}
=11
$