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

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

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

We move all terms to the left:

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


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

-2c(-2c+6)

We multiply parentheses

4c^2-12c

Back to the equation:

-(4c^2-12c)


We get rid of parentheses

-4c^2+3c+12c+2=0

We add all the numbers together, and all the variables

-4c^2+15c+2=0

a = -4; b = 15; c = +2;
Δ = b2-4ac
Δ = 152-4·(-4)·2
Δ = 257
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{257}}{2*-4}=\frac{-15-\sqrt{257}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{257}}{2*-4}=\frac{-15+\sqrt{257}}{-8}
$