-3z^2+26z-35=0

Solution for -3z^2+26z-35=0 equation:

-3z^2+26z-35=0

a = -3; b = 26; c = -35;
Δ = b2-4ac
Δ = 262-4·(-3)·(-35)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-16}{2*-3}=\frac{-42}{-6}
=+7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+16}{2*-3}=\frac{-10}{-6}
=1+2/3
$