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x^2+19.x+90=0
a = 1; b = 19.; c = +90;
Δ = b2-4ac
Δ = 19.2-4·1·90
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19.)-1}{2*1}=\frac{-20}{2} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19.)+1}{2*1}=\frac{-18}{2} =-9 $
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