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8x^2=40
We move all terms to the left:
8x^2-(40)=0
a = 8; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·8·(-40)
Δ = 1280
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{5}}{2*8}=\frac{0-16\sqrt{5}}{16} =-\frac{16\sqrt{5}}{16} =-\sqrt{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{5}}{2*8}=\frac{0+16\sqrt{5}}{16} =\frac{16\sqrt{5}}{16} =\sqrt{5} $
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