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12x^2+13x-14=0
a = 12; b = 13; c = -14;
Δ = b2-4ac
Δ = 132-4·12·(-14)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-29}{2*12}=\frac{-42}{24} =-1+3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+29}{2*12}=\frac{16}{24} =2/3 $
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