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12r^2-13r-35=0
a = 12; b = -13; c = -35;
Δ = b2-4ac
Δ = -132-4·12·(-35)
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$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-43}{2*12}=\frac{-30}{24} =-1+1/4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+43}{2*12}=\frac{56}{24} =2+1/3 $
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