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14x^2+36x-18=0
a = 14; b = 36; c = -18;
Δ = b2-4ac
Δ = 362-4·14·(-18)
Δ = 2304
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{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-48}{2*14}=\frac{-84}{28} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+48}{2*14}=\frac{12}{28} =3/7 $
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