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4x^2=326
We move all terms to the left:
4x^2-(326)=0
a = 4; b = 0; c = -326;
Δ = b2-4ac
Δ = 02-4·4·(-326)
Δ = 5216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5216}=\sqrt{16*326}=\sqrt{16}*\sqrt{326}=4\sqrt{326}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{326}}{2*4}=\frac{0-4\sqrt{326}}{8} =-\frac{4\sqrt{326}}{8} =-\frac{\sqrt{326}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{326}}{2*4}=\frac{0+4\sqrt{326}}{8} =\frac{4\sqrt{326}}{8} =\frac{\sqrt{326}}{2} $
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