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-3r^2-7r+10=0
a = -3; b = -7; c = +10;
Δ = b2-4ac
Δ = -72-4·(-3)·10
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$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-13}{2*-3}=\frac{-6}{-6} =1 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+13}{2*-3}=\frac{20}{-6} =-3+1/3 $
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