(c+4)(c-3)=0

Solution for (c+4)(c-3)=0 equation:

(c+4)(c-3)=0

We multiply parentheses ..

(+c^2-3c+4c-12)=0

We get rid of parentheses

c^2-3c+4c-12=0

We add all the numbers together, and all the variables

c^2+c-12=0

a = 1; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·1·(-12)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*1}=\frac{-8}{2}
=-4
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*1}=\frac{6}{2}
=3
$