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2x^2-25x+57=0
a = 2; b = -25; c = +57;
Δ = b2-4ac
Δ = -252-4·2·57
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-13}{2*2}=\frac{12}{4} =3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+13}{2*2}=\frac{38}{4} =9+1/2 $
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