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12x^2+3x-9=0
a = 12; b = 3; c = -9;
Δ = b2-4ac
Δ = 32-4·12·(-9)
Δ = 441
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{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*12}=\frac{-24}{24} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*12}=\frac{18}{24} =3/4 $
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