(2z+1)(3z+3)=98

Solution for (2z+1)(3z+3)=98 equation:

(2z+1)(3z+3)=98

We move all terms to the left:

(2z+1)(3z+3)-(98)=0

We multiply parentheses ..

(+6z^2+6z+3z+3)-98=0

We get rid of parentheses

6z^2+6z+3z+3-98=0

We add all the numbers together, and all the variables

6z^2+9z-95=0

a = 6; b = 9; c = -95;
Δ = b2-4ac
Δ = 92-4·6·(-95)
Δ = 2361
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{-(9)-\sqrt{2361}}{2*6}=\frac{-9-\sqrt{2361}}{12}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{2361}}{2*6}=\frac{-9+\sqrt{2361}}{12}
$