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12x^2+4x-96=0
a = 12; b = 4; c = -96;
Δ = b2-4ac
Δ = 42-4·12·(-96)
Δ = 4624
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{4624}=68$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-68}{2*12}=\frac{-72}{24} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+68}{2*12}=\frac{64}{24} =2+2/3 $
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