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2x^2+17x+16=0
a = 2; b = 17; c = +16;
Δ = b2-4ac
Δ = 172-4·2·16
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-\sqrt{161}}{2*2}=\frac{-17-\sqrt{161}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{161}}{2*2}=\frac{-17+\sqrt{161}}{4} $
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