45-2c+4=4(c+5)c

Solution for 45-2c+4=4(c+5)c equation:

45-2c+4=4(c+5)c

We move all terms to the left:

45-2c+4-(4(c+5)c)=0

We add all the numbers together, and all the variables

-2c-(4(c+5)c)+49=0


We calculate terms in parentheses: -(4(c+5)c), so:

4(c+5)c

We multiply parentheses

4c^2+20c

Back to the equation:

-(4c^2+20c)


We get rid of parentheses

-4c^2-2c-20c+49=0

We add all the numbers together, and all the variables

-4c^2-22c+49=0

a = -4; b = -22; c = +49;
Δ = b2-4ac
Δ = -222-4·(-4)·49
Δ = 1268
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:

$\sqrt{\Delta}=\sqrt{1268}=\sqrt{4*317}=\sqrt{4}*\sqrt{317}=2\sqrt{317}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{317}}{2*-4}=\frac{22-2\sqrt{317}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{317}}{2*-4}=\frac{22+2\sqrt{317}}{-8}
$