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15x^2+12x-3=0
a = 15; b = 12; c = -3;
Δ = b2-4ac
Δ = 122-4·15·(-3)
Δ = 324
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{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-18}{2*15}=\frac{-30}{30} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+18}{2*15}=\frac{6}{30} =1/5 $
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