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4x^2=99
We move all terms to the left:
4x^2-(99)=0
a = 4; b = 0; c = -99;
Δ = b2-4ac
Δ = 02-4·4·(-99)
Δ = 1584
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1584}=\sqrt{144*11}=\sqrt{144}*\sqrt{11}=12\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{11}}{2*4}=\frac{0-12\sqrt{11}}{8} =-\frac{12\sqrt{11}}{8} =-\frac{3\sqrt{11}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{11}}{2*4}=\frac{0+12\sqrt{11}}{8} =\frac{12\sqrt{11}}{8} =\frac{3\sqrt{11}}{2} $
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