(z-4)(z+3)=60

Solution for (z-4)(z+3)=60 equation:

(z-4)(z+3)=60

We move all terms to the left:

(z-4)(z+3)-(60)=0

We multiply parentheses ..

(+z^2+3z-4z-12)-60=0

We get rid of parentheses

z^2+3z-4z-12-60=0

We add all the numbers together, and all the variables

z^2-1z-72=0

a = 1; b = -1; c = -72;
Δ = b2-4ac
Δ = -12-4·1·(-72)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-17}{2*1}=\frac{-16}{2}
=-8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+17}{2*1}=\frac{18}{2}
=9
$