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4r^2-13r-66=0
a = 4; b = -13; c = -66;
Δ = b2-4ac
Δ = -132-4·4·(-66)
Δ = 1225
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1225}=35$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-35}{2*4}=\frac{-22}{8} =-2+3/4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+35}{2*4}=\frac{48}{8} =6 $
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