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12x^2+3x-6=0
a = 12; b = 3; c = -6;
Δ = b2-4ac
Δ = 32-4·12·(-6)
Δ = 297
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{33}}{2*12}=\frac{-3-3\sqrt{33}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{33}}{2*12}=\frac{-3+3\sqrt{33}}{24} $
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