6z^2+25z-25=0

Solution for 6z^2+25z-25=0 equation:

6z^2+25z-25=0

a = 6; b = 25; c = -25;
Δ = b2-4ac
Δ = 252-4·6·(-25)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-35}{2*6}=\frac{-60}{12}
=-5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+35}{2*6}=\frac{10}{12}
=5/6
$