20=4(c+1)c=

Solution for 20=4(c+1)c= equation:

20=4(c+1)c=

We move all terms to the left:

20-(4(c+1)c)=0


We calculate terms in parentheses: -(4(c+1)c), so:

4(c+1)c

We multiply parentheses

4c^2+4c

Back to the equation:

-(4c^2+4c)


We get rid of parentheses

-4c^2-4c+20=0

a = -4; b = -4; c = +20;
Δ = b2-4ac
Δ = -42-4·(-4)·20
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{21}}{2*-4}=\frac{4-4\sqrt{21}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{21}}{2*-4}=\frac{4+4\sqrt{21}}{-8}
$