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15x^2+18x-168=0
a = 15; b = 18; c = -168;
Δ = b2-4ac
Δ = 182-4·15·(-168)
Δ = 10404
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{10404}=102$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-102}{2*15}=\frac{-120}{30} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+102}{2*15}=\frac{84}{30} =2+4/5 $
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