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

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

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

We move all terms to the left:

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

We multiply parentheses

5c^2-15c-(-(c+2))=0


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

-(c+2)

We get rid of parentheses

-c-2

We add all the numbers together, and all the variables

-1c-2

Back to the equation:

-(-1c-2)


We get rid of parentheses

5c^2-15c+1c+2=0

We add all the numbers together, and all the variables

5c^2-14c+2=0

a = 5; b = -14; c = +2;
Δ = b2-4ac
Δ = -142-4·5·2
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{39}}{2*5}=\frac{14-2\sqrt{39}}{10}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{39}}{2*5}=\frac{14+2\sqrt{39}}{10}
$