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15b^2-32b-7=0
a = 15; b = -32; c = -7;
Δ = b2-4ac
Δ = -322-4·15·(-7)
Δ = 1444
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{1444}=38$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-38}{2*15}=\frac{-6}{30} =-1/5 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+38}{2*15}=\frac{70}{30} =2+1/3 $
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