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4u^2-4u-15=0
a = 4; b = -4; c = -15;
Δ = b2-4ac
Δ = -42-4·4·(-15)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-16}{2*4}=\frac{-12}{8} =-1+1/2 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+16}{2*4}=\frac{20}{8} =2+1/2 $
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