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

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## Solution for (5r-3)(r-8)=0 equation:

(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 = b2-4acΔ = -432-4·5·24Δ = 1369The delta value is higher than zero, so the equation has two solutionsWe 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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