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12r^2-9r-5=0
a = 12; b = -9; c = -5;
Δ = b2-4ac
Δ = -92-4·12·(-5)
Δ = 321
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{321}}{2*12}=\frac{9-\sqrt{321}}{24} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{321}}{2*12}=\frac{9+\sqrt{321}}{24} $
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