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15b^2+4b-4=0
a = 15; b = 4; c = -4;
Δ = b2-4ac
Δ = 42-4·15·(-4)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*15}=\frac{-20}{30} =-2/3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*15}=\frac{12}{30} =2/5 $
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