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2r^2+11r+15=0
a = 2; b = 11; c = +15;
Δ = b2-4ac
Δ = 112-4·2·15
Δ = 1
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{1}=1$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-1}{2*2}=\frac{-12}{4} =-3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*2}=\frac{-10}{4} =-2+1/2 $
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