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11u-15-2u^2=0
a = -2; b = 11; c = -15;
Δ = b2-4ac
Δ = 112-4·(-2)·(-15)
Δ = 1
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{1}=1$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-1}{2*-2}=\frac{-12}{-4} =+3 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*-2}=\frac{-10}{-4} =2+1/2 $
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