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100=-5(r-7)8r
We move all terms to the left:
100-(-5(r-7)8r)=0
We calculate terms in parentheses: -(-5(r-7)8r), so:We get rid of parentheses
-5(r-7)8r
We multiply parentheses
-40r^2+280r
Back to the equation:
-(-40r^2+280r)
40r^2-280r+100=0
a = 40; b = -280; c = +100;
Δ = b2-4ac
Δ = -2802-4·40·100
Δ = 62400
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{62400}=\sqrt{1600*39}=\sqrt{1600}*\sqrt{39}=40\sqrt{39}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-280)-40\sqrt{39}}{2*40}=\frac{280-40\sqrt{39}}{80} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-280)+40\sqrt{39}}{2*40}=\frac{280+40\sqrt{39}}{80} $
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