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2x^2-24x+2=0
a = 2; b = -24; c = +2;
Δ = b2-4ac
Δ = -242-4·2·2
Δ = 560
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{35}}{2*2}=\frac{24-4\sqrt{35}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{35}}{2*2}=\frac{24+4\sqrt{35}}{4} $
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