20z^2+13z-15=0

Solution for 20z^2+13z-15=0 equation:

20z^2+13z-15=0

a = 20; b = 13; c = -15;
Δ = b2-4ac
Δ = 132-4·20·(-15)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-37}{2*20}=\frac{-50}{40}
=-1+1/4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+37}{2*20}=\frac{24}{40}
=3/5
$