51-(2c+3)=4(c+5)c

Solution for 51-(2c+3)=4(c+5)c equation:

51-(2c+3)=4(c+5)c

We move all terms to the left:

51-(2c+3)-(4(c+5)c)=0

We get rid of parentheses

-2c-(4(c+5)c)-3+51=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 add all the numbers together, and all the variables

-2c-(4c^2+20c)+48=0

We get rid of parentheses

-4c^2-2c-20c+48=0

We add all the numbers together, and all the variables

-4c^2-22c+48=0

a = -4; b = -22; c = +48;
Δ = b2-4ac
Δ = -222-4·(-4)·48
Δ = 1252
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:

$\sqrt{\Delta}=\sqrt{1252}=\sqrt{4*313}=\sqrt{4}*\sqrt{313}=2\sqrt{313}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{313}}{2*-4}=\frac{22-2\sqrt{313}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{313}}{2*-4}=\frac{22+2\sqrt{313}}{-8}
$