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2x^2-45x+250=0
a = 2; b = -45; c = +250;
Δ = b2-4ac
Δ = -452-4·2·250
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-5}{2*2}=\frac{40}{4} =10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+5}{2*2}=\frac{50}{4} =12+1/2 $
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