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5z^2+16z+3=0
a = 5; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·5·3
Δ = 196
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{196}=14$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-14}{2*5}=\frac{-30}{10} =-3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14}{2*5}=\frac{-2}{10} =-1/5 $
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