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2x^2-24x=-48
We move all terms to the left:
2x^2-24x-(-48)=0
We add all the numbers together, and all the variables
2x^2-24x+48=0
a = 2; b = -24; c = +48;
Δ = b2-4ac
Δ = -242-4·2·48
Δ = 192
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8\sqrt{3}}{2*2}=\frac{24-8\sqrt{3}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8\sqrt{3}}{2*2}=\frac{24+8\sqrt{3}}{4} $
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