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3x^2+16x+21=0
a = 3; b = 16; c = +21;
Δ = b2-4ac
Δ = 162-4·3·21
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2}{2*3}=\frac{-18}{6} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2}{2*3}=\frac{-14}{6} =-2+1/3 $
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