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12x^2-22x-4=0
a = 12; b = -22; c = -4;
Δ = b2-4ac
Δ = -222-4·12·(-4)
Δ = 676
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{676}=26$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-26}{2*12}=\frac{-4}{24} =-1/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+26}{2*12}=\frac{48}{24} =2 $
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