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5r^2+10r-15=0
a = 5; b = 10; c = -15;
Δ = b2-4ac
Δ = 102-4·5·(-15)
Δ = 400
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{400}=20$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-20}{2*5}=\frac{-30}{10} =-3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+20}{2*5}=\frac{10}{10} =1 $
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