(c)=3c^2+4c-5

Solution for (c)=3c^2+4c-5 equation:

(c)=3c^2+4c-5

We move all terms to the left:

(c)-(3c^2+4c-5)=0

We get rid of parentheses

-3c^2+c-4c+5=0

We add all the numbers together, and all the variables

-3c^2-3c+5=0

a = -3; b = -3; c = +5;
Δ = b2-4ac
Δ = -32-4·(-3)·5
Δ = 69
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{-(-3)-\sqrt{69}}{2*-3}=\frac{3-\sqrt{69}}{-6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{69}}{2*-3}=\frac{3+\sqrt{69}}{-6}
$