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20-5x-4x^2=0
a = -4; b = -5; c = +20;
Δ = b2-4ac
Δ = -52-4·(-4)·20
Δ = 345
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{345}}{2*-4}=\frac{5-\sqrt{345}}{-8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{345}}{2*-4}=\frac{5+\sqrt{345}}{-8} $
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