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12x^2+4x-40=0
a = 12; b = 4; c = -40;
Δ = b2-4ac
Δ = 42-4·12·(-40)
Δ = 1936
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{1936}=44$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-44}{2*12}=\frac{-48}{24} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+44}{2*12}=\frac{40}{24} =1+2/3 $
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