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5r^2+21r+4=0
a = 5; b = 21; c = +4;
Δ = b2-4ac
Δ = 212-4·5·4
Δ = 361
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{361}=19$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-19}{2*5}=\frac{-40}{10} =-4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+19}{2*5}=\frac{-2}{10} =-1/5 $
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