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2x^2+18x+16=0
a = 2; b = 18; c = +16;
Δ = b2-4ac
Δ = 182-4·2·16
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-14}{2*2}=\frac{-32}{4} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+14}{2*2}=\frac{-4}{4} =-1 $
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