35z^2-26z-48=0

Solution for 35z^2-26z-48=0 equation:

35z^2-26z-48=0

a = 35; b = -26; c = -48;
Δ = b2-4ac
Δ = -262-4·35·(-48)
Δ = 7396
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7396}=86$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-86}{2*35}=\frac{-60}{70}
=-6/7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+86}{2*35}=\frac{112}{70}
=1+3/5
$