4u^2+24u+29=1

Solution for 4u^2+24u+29=1 equation:

4u^2+24u+29=1

We move all terms to the left:

4u^2+24u+29-(1)=0

We add all the numbers together, and all the variables

4u^2+24u+28=0

a = 4; b = 24; c = +28;
Δ = b2-4ac
Δ = 242-4·4·28
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{2}}{2*4}=\frac{-24-8\sqrt{2}}{8}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{2}}{2*4}=\frac{-24+8\sqrt{2}}{8}
$