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2x^2-14x=26
We move all terms to the left:
2x^2-14x-(26)=0
a = 2; b = -14; c = -26;
Δ = b2-4ac
Δ = -142-4·2·(-26)
Δ = 404
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{404}=\sqrt{4*101}=\sqrt{4}*\sqrt{101}=2\sqrt{101}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{101}}{2*2}=\frac{14-2\sqrt{101}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{101}}{2*2}=\frac{14+2\sqrt{101}}{4} $
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