15z^2+17z+4=0

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Solution for 15z^2+17z+4=0 equation:



15z^2+17z+4=0
a = 15; b = 17; c = +4;
Δ = b2-4ac
Δ = 172-4·15·4
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-7}{2*15}=\frac{-24}{30} =-4/5 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+7}{2*15}=\frac{-10}{30} =-1/3 $

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