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4b^2-15b-2=0
a = 4; b = -15; c = -2;
Δ = b2-4ac
Δ = -152-4·4·(-2)
Δ = 257
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{257}}{2*4}=\frac{15-\sqrt{257}}{8} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{257}}{2*4}=\frac{15+\sqrt{257}}{8} $
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