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3x^2-2(4)x+(4)=0
a = 3; b = -24; c = +4;
Δ = b2-4ac
Δ = -242-4·3·4
Δ = 528
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{33}}{2*3}=\frac{24-4\sqrt{33}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{33}}{2*3}=\frac{24+4\sqrt{33}}{6} $
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