-3z^2-9z+32=4z+25

Solution for -3z^2-9z+32=4z+25 equation:

-3z^2-9z+32=4z+25

We move all terms to the left:

-3z^2-9z+32-(4z+25)=0

We get rid of parentheses

-3z^2-9z-4z-25+32=0

We add all the numbers together, and all the variables

-3z^2-13z+7=0

a = -3; b = -13; c = +7;
Δ = b2-4ac
Δ = -132-4·(-3)·7
Δ = 253
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{-(-13)-\sqrt{253}}{2*-3}=\frac{13-\sqrt{253}}{-6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{253}}{2*-3}=\frac{13+\sqrt{253}}{-6}
$