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x2-45=4050
We move all terms to the left:
x2-45-(4050)=0
We add all the numbers together, and all the variables
x^2-4095=0
a = 1; b = 0; c = -4095;
Δ = b2-4ac
Δ = 02-4·1·(-4095)
Δ = 16380
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{16380}=\sqrt{36*455}=\sqrt{36}*\sqrt{455}=6\sqrt{455}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{455}}{2*1}=\frac{0-6\sqrt{455}}{2} =-\frac{6\sqrt{455}}{2} =-3\sqrt{455} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{455}}{2*1}=\frac{0+6\sqrt{455}}{2} =\frac{6\sqrt{455}}{2} =3\sqrt{455} $
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