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4x^2+18x+20=0
a = 4; b = 18; c = +20;
Δ = b2-4ac
Δ = 182-4·4·20
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2}{2*4}=\frac{-20}{8} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2}{2*4}=\frac{-16}{8} =-2 $
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