21z^2+5z-26=0

Solution for 21z^2+5z-26=0 equation:

21z^2+5z-26=0

a = 21; b = 5; c = -26;
Δ = b2-4ac
Δ = 52-4·21·(-26)
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-47}{2*21}=\frac{-52}{42}
=-1+5/21
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+47}{2*21}=\frac{42}{42}
=1
$