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5x^2-19x+18=0
a = 5; b = -19; c = +18;
Δ = b2-4ac
Δ = -192-4·5·18
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-1}{2*5}=\frac{18}{10} =1+4/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+1}{2*5}=\frac{20}{10} =2 $
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