25-(2c+3)=2(c+6)c

Solution for 25-(2c+3)=2(c+6)c equation:

25-(2c+3)=2(c+6)c

We move all terms to the left:

25-(2c+3)-(2(c+6)c)=0

We get rid of parentheses

-2c-(2(c+6)c)-3+25=0


We calculate terms in parentheses: -(2(c+6)c), so:

2(c+6)c

We multiply parentheses

2c^2+12c

Back to the equation:

-(2c^2+12c)


We add all the numbers together, and all the variables

-2c-(2c^2+12c)+22=0

We get rid of parentheses

-2c^2-2c-12c+22=0

We add all the numbers together, and all the variables

-2c^2-14c+22=0

a = -2; b = -14; c = +22;
Δ = b2-4ac
Δ = -142-4·(-2)·22
Δ = 372
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:

$\sqrt{\Delta}=\sqrt{372}=\sqrt{4*93}=\sqrt{4}*\sqrt{93}=2\sqrt{93}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{93}}{2*-2}=\frac{14-2\sqrt{93}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{93}}{2*-2}=\frac{14+2\sqrt{93}}{-4}
$