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16x^2=300
We move all terms to the left:
16x^2-(300)=0
a = 16; b = 0; c = -300;
Δ = b2-4ac
Δ = 02-4·16·(-300)
Δ = 19200
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{19200}=\sqrt{6400*3}=\sqrt{6400}*\sqrt{3}=80\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80\sqrt{3}}{2*16}=\frac{0-80\sqrt{3}}{32} =-\frac{80\sqrt{3}}{32} =-\frac{5\sqrt{3}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80\sqrt{3}}{2*16}=\frac{0+80\sqrt{3}}{32} =\frac{80\sqrt{3}}{32} =\frac{5\sqrt{3}}{2} $
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