4(c+2)4c+10=1

Solution for 4(c+2)4c+10=1 equation:

4(c+2)4c+10=1

We move all terms to the left:

4(c+2)4c+10-(1)=0

We add all the numbers together, and all the variables

4(c+2)4c+9=0

We multiply parentheses

16c^2+32c+9=0

a = 16; b = 32; c = +9;
Δ = b2-4ac
Δ = 322-4·16·9
Δ = 448
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{7}}{2*16}=\frac{-32-8\sqrt{7}}{32}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{7}}{2*16}=\frac{-32+8\sqrt{7}}{32}
$