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15x^2+28x-4=0
a = 15; b = 28; c = -4;
Δ = b2-4ac
Δ = 282-4·15·(-4)
Δ = 1024
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{1024}=32$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-32}{2*15}=\frac{-60}{30} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+32}{2*15}=\frac{4}{30} =2/15 $
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