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10x^2=16
We move all terms to the left:
10x^2-(16)=0
a = 10; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·10·(-16)
Δ = 640
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{10}}{2*10}=\frac{0-8\sqrt{10}}{20} =-\frac{8\sqrt{10}}{20} =-\frac{2\sqrt{10}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{10}}{2*10}=\frac{0+8\sqrt{10}}{20} =\frac{8\sqrt{10}}{20} =\frac{2\sqrt{10}}{5} $
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