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2x^2+43x-399=0
a = 2; b = 43; c = -399;
Δ = b2-4ac
Δ = 432-4·2·(-399)
Δ = 5041
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{5041}=71$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-71}{2*2}=\frac{-114}{4} =-28+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+71}{2*2}=\frac{28}{4} =7 $
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