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2r^2+5r-18=0
a = 2; b = 5; c = -18;
Δ = b2-4ac
Δ = 52-4·2·(-18)
Δ = 169
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{169}=13$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*2}=\frac{-18}{4} =-4+1/2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*2}=\frac{8}{4} =2 $
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