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12a^2+11a+2=0
a = 12; b = 11; c = +2;
Δ = b2-4ac
Δ = 112-4·12·2
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*12}=\frac{-16}{24} =-2/3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*12}=\frac{-6}{24} =-1/4 $
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