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1+12x-3x^2=0
a = -3; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·(-3)·1
Δ = 156
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{39}}{2*-3}=\frac{-12-2\sqrt{39}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{39}}{2*-3}=\frac{-12+2\sqrt{39}}{-6} $
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