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(5r-3)(r-8)=0

We multiply parentheses ..

(+5r^2-40r-3r+24)=0

We get rid of parentheses

5r^2-40r-3r+24=0

We add all the numbers together, and all the variables

5r^2-43r+24=0

a = 5; b = -43; c = +24;

Δ = b^{2}-4ac

Δ = -43^{2}-4·5·24

Δ = 1369

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{1369}=37$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-43)-37}{2*5}=\frac{6}{10} =3/5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-43)+37}{2*5}=\frac{80}{10} =8 $

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