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2x^2-46x+252=0
a = 2; b = -46; c = +252;
Δ = b2-4ac
Δ = -462-4·2·252
Δ = 100
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{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-46)-10}{2*2}=\frac{36}{4} =9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-46)+10}{2*2}=\frac{56}{4} =14 $
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