-42=5(2c+9)-3c(c+8)

Solution for -42=5(2c+9)-3c(c+8) equation:

-42=5(2c+9)-3c(c+8)

We move all terms to the left:

-42-(5(2c+9)-3c(c+8))=0


We calculate terms in parentheses: -(5(2c+9)-3c(c+8)), so:

5(2c+9)-3c(c+8)

We multiply parentheses

-3c^2+10c-24c+45

We add all the numbers together, and all the variables

-3c^2-14c+45

Back to the equation:

-(-3c^2-14c+45)


We get rid of parentheses

3c^2+14c-45-42=0

We add all the numbers together, and all the variables

3c^2+14c-87=0

a = 3; b = 14; c = -87;
Δ = b2-4ac
Δ = 142-4·3·(-87)
Δ = 1240
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:

$\sqrt{\Delta}=\sqrt{1240}=\sqrt{4*310}=\sqrt{4}*\sqrt{310}=2\sqrt{310}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{310}}{2*3}=\frac{-14-2\sqrt{310}}{6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{310}}{2*3}=\frac{-14+2\sqrt{310}}{6}
$