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12x^2-36x+1.5=0
a = 12; b = -36; c = +1.5;
Δ = b2-4ac
Δ = -362-4·12·1.5
Δ = 1224
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1224}=\sqrt{36*34}=\sqrt{36}*\sqrt{34}=6\sqrt{34}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-6\sqrt{34}}{2*12}=\frac{36-6\sqrt{34}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+6\sqrt{34}}{2*12}=\frac{36+6\sqrt{34}}{24} $
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