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4x^2-48=0
a = 4; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·4·(-48)
Δ = 768
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*4}=\frac{0-16\sqrt{3}}{8} =-\frac{16\sqrt{3}}{8} =-2\sqrt{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*4}=\frac{0+16\sqrt{3}}{8} =\frac{16\sqrt{3}}{8} =2\sqrt{3} $
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