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2x^2+19x+45=0
a = 2; b = 19; c = +45;
Δ = b2-4ac
Δ = 192-4·2·45
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-1}{2*2}=\frac{-20}{4} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+1}{2*2}=\frac{-18}{4} =-4+1/2 $
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