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12x^2+24x-16=0
a = 12; b = 24; c = -16;
Δ = b2-4ac
Δ = 242-4·12·(-16)
Δ = 1344
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1344}=\sqrt{64*21}=\sqrt{64}*\sqrt{21}=8\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{21}}{2*12}=\frac{-24-8\sqrt{21}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{21}}{2*12}=\frac{-24+8\sqrt{21}}{24} $
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