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2x^2-35x+98=0
a = 2; b = -35; c = +98;
Δ = b2-4ac
Δ = -352-4·2·98
Δ = 441
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{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-21}{2*2}=\frac{14}{4} =3+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+21}{2*2}=\frac{56}{4} =14 $
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