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15x^2+42x-9=0
a = 15; b = 42; c = -9;
Δ = b2-4ac
Δ = 422-4·15·(-9)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-48}{2*15}=\frac{-90}{30} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+48}{2*15}=\frac{6}{30} =1/5 $
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