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2c^2+17c+36=0
a = 2; b = 17; c = +36;
Δ = b2-4ac
Δ = 172-4·2·36
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-1}{2*2}=\frac{-18}{4} =-4+1/2 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+1}{2*2}=\frac{-16}{4} =-4 $
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