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10x^2+42x+44=0
a = 10; b = 42; c = +44;
Δ = b2-4ac
Δ = 422-4·10·44
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2}{2*10}=\frac{-44}{20} =-2+1/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2}{2*10}=\frac{-40}{20} =-2 $
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