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11x^2+28x+12=0
a = 11; b = 28; c = +12;
Δ = b2-4ac
Δ = 282-4·11·12
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-16}{2*11}=\frac{-44}{22} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+16}{2*11}=\frac{-12}{22} =-6/11 $
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