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2+5r-10r^2=0
a = -10; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·(-10)·2
Δ = 105
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{105}}{2*-10}=\frac{-5-\sqrt{105}}{-20} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{105}}{2*-10}=\frac{-5+\sqrt{105}}{-20} $
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