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4b^2-3b-15=0
a = 4; b = -3; c = -15;
Δ = b2-4ac
Δ = -32-4·4·(-15)
Δ = 249
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{-(-3)-\sqrt{249}}{2*4}=\frac{3-\sqrt{249}}{8} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{249}}{2*4}=\frac{3+\sqrt{249}}{8} $
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