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-12r^2+5r+3=0
a = -12; b = 5; c = +3;
Δ = b2-4ac
Δ = 52-4·(-12)·3
Δ = 169
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{169}=13$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*-12}=\frac{-18}{-24} =3/4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*-12}=\frac{8}{-24} =-1/3 $
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