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2x^2-24x-4=0
a = 2; b = -24; c = -4;
Δ = b2-4ac
Δ = -242-4·2·(-4)
Δ = 608
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{608}=\sqrt{16*38}=\sqrt{16}*\sqrt{38}=4\sqrt{38}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{38}}{2*2}=\frac{24-4\sqrt{38}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{38}}{2*2}=\frac{24+4\sqrt{38}}{4} $
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