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8a^2-12a+3=0
a = 8; b = -12; c = +3;
Δ = b2-4ac
Δ = -122-4·8·3
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{3}}{2*8}=\frac{12-4\sqrt{3}}{16} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{3}}{2*8}=\frac{12+4\sqrt{3}}{16} $
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