4x^2-20x^2+25=2

Solution for 4x^2-20x^2+25=2 equation:

4x^2-20x^2+25=2

We move all terms to the left:

4x^2-20x^2+25-(2)=0

We add all the numbers together, and all the variables

-16x^2+23=0

a = -16; b = 0; c = +23;
Δ = b2-4ac
Δ = 02-4·(-16)·23
Δ = 1472
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:

$\sqrt{\Delta}=\sqrt{1472}=\sqrt{64*23}=\sqrt{64}*\sqrt{23}=8\sqrt{23}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{23}}{2*-16}=\frac{0-8\sqrt{23}}{-32}
=-\frac{8\sqrt{23}}{-32}
=-\frac{\sqrt{23}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{23}}{2*-16}=\frac{0+8\sqrt{23}}{-32}
=\frac{8\sqrt{23}}{-32}
=\frac{\sqrt{23}}{-4}
$