4x^2+1=7

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Solution for 4x^2+1=7 equation:



4x^2+1=7
We move all terms to the left:
4x^2+1-(7)=0
We add all the numbers together, and all the variables
4x^2-6=0
a = 4; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·4·(-6)
Δ = 96
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:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*4}=\frac{0-4\sqrt{6}}{8} =-\frac{4\sqrt{6}}{8} =-\frac{\sqrt{6}}{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*4}=\frac{0+4\sqrt{6}}{8} =\frac{4\sqrt{6}}{8} =\frac{\sqrt{6}}{2} $

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