4x^2+9/5=5

Solution for 4x^2+9/5=5 equation:

4x^2+9/5=5

We move all terms to the left:

4x^2+9/5-(5)=0

determiningTheFunctionDomain
4x^2-5+9/5=0

We multiply all the terms by the denominator


4x^2*5+9-5*5=0

We add all the numbers together, and all the variables

4x^2*5-16=0

Wy multiply elements

20x^2-16=0

a = 20; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·20·(-16)
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{5}}{2*20}=\frac{0-16\sqrt{5}}{40}
=-\frac{16\sqrt{5}}{40}
=-\frac{2\sqrt{5}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{5}}{2*20}=\frac{0+16\sqrt{5}}{40}
=\frac{16\sqrt{5}}{40}
=\frac{2\sqrt{5}}{5}
$