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2x^2-18=8x
We move all terms to the left:
2x^2-18-(8x)=0
a = 2; b = -8; c = -18;
Δ = b2-4ac
Δ = -82-4·2·(-18)
Δ = 208
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{13}}{2*2}=\frac{8-4\sqrt{13}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{13}}{2*2}=\frac{8+4\sqrt{13}}{4} $
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