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2x^2+21x+55=6
We move all terms to the left:
2x^2+21x+55-(6)=0
We add all the numbers together, and all the variables
2x^2+21x+49=0
a = 2; b = 21; c = +49;
Δ = b2-4ac
Δ = 212-4·2·49
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-7}{2*2}=\frac{-28}{4} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+7}{2*2}=\frac{-14}{4} =-3+1/2 $
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