(z^2+3)+(4z-7)-(5z^2+z-9)=0

Solution for (z^2+3)+(4z-7)-(5z^2+z-9)=0 equation:

(z^2+3)+(4z-7)-(5z^2+z-9)=0

We get rid of parentheses

z^2-5z^2+4z-z+3-7+9=0

We add all the numbers together, and all the variables

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

a = -4; b = 3; c = +5;
Δ = b2-4ac
Δ = 32-4·(-4)·5
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{89}}{2*-4}=\frac{-3-\sqrt{89}}{-8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{89}}{2*-4}=\frac{-3+\sqrt{89}}{-8}
$