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2x^2=-19x-9
We move all terms to the left:
2x^2-(-19x-9)=0
We get rid of parentheses
2x^2+19x+9=0
a = 2; b = 19; c = +9;
Δ = b2-4ac
Δ = 192-4·2·9
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-17}{2*2}=\frac{-36}{4} =-9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+17}{2*2}=\frac{-2}{4} =-1/2 $
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