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(7r-5)(r+5)=0
We multiply parentheses ..
(+7r^2+35r-5r-25)=0
We get rid of parentheses
7r^2+35r-5r-25=0
We add all the numbers together, and all the variables
7r^2+30r-25=0
a = 7; b = 30; c = -25;
Δ = b2-4ac
Δ = 302-4·7·(-25)
Δ = 1600
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{1600}=40$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-40}{2*7}=\frac{-70}{14} =-5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+40}{2*7}=\frac{10}{14} =5/7 $
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