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4r^2-4r-3=0
a = 4; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·4·(-3)
Δ = 64
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{64}=8$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*4}=\frac{-4}{8} =-1/2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*4}=\frac{12}{8} =1+1/2 $
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