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16x^2=320
We move all terms to the left:
16x^2-(320)=0
a = 16; b = 0; c = -320;
Δ = b2-4ac
Δ = 02-4·16·(-320)
Δ = 20480
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20480}=\sqrt{4096*5}=\sqrt{4096}*\sqrt{5}=64\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64\sqrt{5}}{2*16}=\frac{0-64\sqrt{5}}{32} =-\frac{64\sqrt{5}}{32} =-2\sqrt{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64\sqrt{5}}{2*16}=\frac{0+64\sqrt{5}}{32} =\frac{64\sqrt{5}}{32} =2\sqrt{5} $
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