5c^2+4c=+1-c+8

Solution for 5c^2+4c=+1-c+8 equation:

5c^2+4c=+1-c+8

We move all terms to the left:

5c^2+4c-(+1-c+8)=0

We add all the numbers together, and all the variables

5c^2+4c-(-1c+9)=0

We get rid of parentheses

5c^2+4c+1c-9=0

We add all the numbers together, and all the variables

5c^2+5c-9=0

a = 5; b = 5; c = -9;
Δ = b2-4ac
Δ = 52-4·5·(-9)
Δ = 205
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{-(5)-\sqrt{205}}{2*5}=\frac{-5-\sqrt{205}}{10}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{205}}{2*5}=\frac{-5+\sqrt{205}}{10}
$