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(3x)^2+19x+16=0
a = 3; b = 19; c = +16;
Δ = b2-4ac
Δ = 192-4·3·16
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*3}=\frac{-32}{6} =-5+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*3}=\frac{-6}{6} =-1 $
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