2u^2+8u+24=u^2+12u+36

Solution for 2u^2+8u+24=u^2+12u+36 equation:

2u^2+8u+24=u^2+12u+36

We move all terms to the left:

2u^2+8u+24-(u^2+12u+36)=0

We get rid of parentheses

2u^2-u^2+8u-12u-36+24=0

We add all the numbers together, and all the variables

u^2-4u-12=0

a = 1; b = -4; c = -12;
Δ = b2-4ac
Δ = -42-4·1·(-12)
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{-(-4)-8}{2*1}=\frac{-4}{2}
=-2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*1}=\frac{12}{2}
=6
$