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36z^2+96z+15=0
a = 36; b = 96; c = +15;
Δ = b2-4ac
Δ = 962-4·36·15
Δ = 7056
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{7056}=84$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-84}{2*36}=\frac{-180}{72} =-2+1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+84}{2*36}=\frac{-12}{72} =-1/6 $
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