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4r^2-9r-29=0
a = 4; b = -9; c = -29;
Δ = b2-4ac
Δ = -92-4·4·(-29)
Δ = 545
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{-(-9)-\sqrt{545}}{2*4}=\frac{9-\sqrt{545}}{8} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{545}}{2*4}=\frac{9+\sqrt{545}}{8} $
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