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20x^2+8x-64=0
a = 20; b = 8; c = -64;
Δ = b2-4ac
Δ = 82-4·20·(-64)
Δ = 5184
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{5184}=72$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-72}{2*20}=\frac{-80}{40} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+72}{2*20}=\frac{64}{40} =1+3/5 $
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