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20x^2+53x+18=0
a = 20; b = 53; c = +18;
Δ = b2-4ac
Δ = 532-4·20·18
Δ = 1369
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{1369}=37$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(53)-37}{2*20}=\frac{-90}{40} =-2+1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53)+37}{2*20}=\frac{-16}{40} =-2/5 $
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