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4x^2=320
We move all terms to the left:
4x^2-(320)=0
a = 4; b = 0; c = -320;
Δ = b2-4ac
Δ = 02-4·4·(-320)
Δ = 5120
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5120}=\sqrt{1024*5}=\sqrt{1024}*\sqrt{5}=32\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{5}}{2*4}=\frac{0-32\sqrt{5}}{8} =-\frac{32\sqrt{5}}{8} =-4\sqrt{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{5}}{2*4}=\frac{0+32\sqrt{5}}{8} =\frac{32\sqrt{5}}{8} =4\sqrt{5} $
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