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3q^2+16q+5=0
a = 3; b = 16; c = +5;
Δ = b2-4ac
Δ = 162-4·3·5
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-14}{2*3}=\frac{-30}{6} =-5 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14}{2*3}=\frac{-2}{6} =-1/3 $
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