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20x^2-25x-4=0
a = 20; b = -25; c = -4;
Δ = b2-4ac
Δ = -252-4·20·(-4)
Δ = 945
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{945}=\sqrt{9*105}=\sqrt{9}*\sqrt{105}=3\sqrt{105}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-3\sqrt{105}}{2*20}=\frac{25-3\sqrt{105}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+3\sqrt{105}}{2*20}=\frac{25+3\sqrt{105}}{40} $
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