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2u^2+13u-15=0
a = 2; b = 13; c = -15;
Δ = b2-4ac
Δ = 132-4·2·(-15)
Δ = 289
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{289}=17$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*2}=\frac{-30}{4} =-7+1/2 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*2}=\frac{4}{4} =1 $
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