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12x^2+3x-16=0
a = 12; b = 3; c = -16;
Δ = b2-4ac
Δ = 32-4·12·(-16)
Δ = 777
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{777}}{2*12}=\frac{-3-\sqrt{777}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{777}}{2*12}=\frac{-3+\sqrt{777}}{24} $
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