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2x^2+36x-38=0
a = 2; b = 36; c = -38;
Δ = b2-4ac
Δ = 362-4·2·(-38)
Δ = 1600
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{1600}=40$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-40}{2*2}=\frac{-76}{4} =-19 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+40}{2*2}=\frac{4}{4} =1 $
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