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16x^2+42x-18=0
a = 16; b = 42; c = -18;
Δ = b2-4ac
Δ = 422-4·16·(-18)
Δ = 2916
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{2916}=54$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-54}{2*16}=\frac{-96}{32} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+54}{2*16}=\frac{12}{32} =3/8 $
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