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4r^2+13r+3=0
a = 4; b = 13; c = +3;
Δ = b2-4ac
Δ = 132-4·4·3
Δ = 121
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{121}=11$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*4}=\frac{-24}{8} =-3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*4}=\frac{-2}{8} =-1/4 $
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