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2x^2-3x-989=0
a = 2; b = -3; c = -989;
Δ = b2-4ac
Δ = -32-4·2·(-989)
Δ = 7921
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{7921}=89$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-89}{2*2}=\frac{-86}{4} =-21+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+89}{2*2}=\frac{92}{4} =23 $
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