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2r^2-3r-16=0
a = 2; b = -3; c = -16;
Δ = b2-4ac
Δ = -32-4·2·(-16)
Δ = 137
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{137}}{2*2}=\frac{3-\sqrt{137}}{4} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{137}}{2*2}=\frac{3+\sqrt{137}}{4} $
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