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2x^2-24x+8=0
a = 2; b = -24; c = +8;
Δ = b2-4ac
Δ = -242-4·2·8
Δ = 512
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-16\sqrt{2}}{2*2}=\frac{24-16\sqrt{2}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+16\sqrt{2}}{2*2}=\frac{24+16\sqrt{2}}{4} $
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