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16x^2+3x-138=0
a = 16; b = 3; c = -138;
Δ = b2-4ac
Δ = 32-4·16·(-138)
Δ = 8841
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{8841}}{2*16}=\frac{-3-\sqrt{8841}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{8841}}{2*16}=\frac{-3+\sqrt{8841}}{32} $
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