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2r^2-10r-28=0
a = 2; b = -10; c = -28;
Δ = b2-4ac
Δ = -102-4·2·(-28)
Δ = 324
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{324}=18$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-18}{2*2}=\frac{-8}{4} =-2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+18}{2*2}=\frac{28}{4} =7 $
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