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4r^2-3r-115=0
a = 4; b = -3; c = -115;
Δ = b2-4ac
Δ = -32-4·4·(-115)
Δ = 1849
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{1849}=43$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-43}{2*4}=\frac{-40}{8} =-5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+43}{2*4}=\frac{46}{8} =5+3/4 $
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