3z^2+26z=9

Solution for 3z^2+26z=9 equation:

3z^2+26z=9

We move all terms to the left:

3z^2+26z-(9)=0

a = 3; b = 26; c = -9;
Δ = b2-4ac
Δ = 262-4·3·(-9)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-28}{2*3}=\frac{-54}{6}
=-9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+28}{2*3}=\frac{2}{6}
=1/3
$