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20x^2=32
We move all terms to the left:
20x^2-(32)=0
a = 20; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·20·(-32)
Δ = 2560
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{10}}{2*20}=\frac{0-16\sqrt{10}}{40} =-\frac{16\sqrt{10}}{40} =-\frac{2\sqrt{10}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{10}}{2*20}=\frac{0+16\sqrt{10}}{40} =\frac{16\sqrt{10}}{40} =\frac{2\sqrt{10}}{5} $
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