9c(c-11)+10(5c-3)=3(c+5)+c(6c-3)-30

Solution for 9c(c-11)+10(5c-3)=3(c+5)+c(6c-3)-30 equation:

9c(c-11)+10(5c-3)=3(c+5)+c(6c-3)-30

We move all terms to the left:

9c(c-11)+10(5c-3)-(3(c+5)+c(6c-3)-30)=0

We multiply parentheses

9c^2-99c+50c-(3(c+5)+c(6c-3)-30)-30=0


We calculate terms in parentheses: -(3(c+5)+c(6c-3)-30), so:

3(c+5)+c(6c-3)-30

We multiply parentheses

6c^2+3c-3c+15-30

We add all the numbers together, and all the variables

6c^2-15

Back to the equation:

-(6c^2-15)


We add all the numbers together, and all the variables

9c^2-49c-(6c^2-15)-30=0

We get rid of parentheses

9c^2-6c^2-49c+15-30=0

We add all the numbers together, and all the variables

3c^2-49c-15=0

a = 3; b = -49; c = -15;
Δ = b2-4ac
Δ = -492-4·3·(-15)
Δ = 2581
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{-(-49)-\sqrt{2581}}{2*3}=\frac{49-\sqrt{2581}}{6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+\sqrt{2581}}{2*3}=\frac{49+\sqrt{2581}}{6}
$