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2x^2-42x+216=0
a = 2; b = -42; c = +216;
Δ = b2-4ac
Δ = -422-4·2·216
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6}{2*2}=\frac{36}{4} =9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6}{2*2}=\frac{48}{4} =12 $
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