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12x^2-4x-5=0
a = 12; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·12·(-5)
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-16}{2*12}=\frac{-12}{24} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+16}{2*12}=\frac{20}{24} =5/6 $
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