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2x^2=-12x+18
We move all terms to the left:
2x^2-(-12x+18)=0
We get rid of parentheses
2x^2+12x-18=0
a = 2; b = 12; c = -18;
Δ = b2-4ac
Δ = 122-4·2·(-18)
Δ = 288
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{2}}{2*2}=\frac{-12-12\sqrt{2}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{2}}{2*2}=\frac{-12+12\sqrt{2}}{4} $
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