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14x^2-13x-12=0
a = 14; b = -13; c = -12;
Δ = b2-4ac
Δ = -132-4·14·(-12)
Δ = 841
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{841}=29$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-29}{2*14}=\frac{-16}{28} =-4/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+29}{2*14}=\frac{42}{28} =1+1/2 $
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