x^2+(25-3x/4)^2=25

Solution for x^2+(25-3x/4)^2=25 equation:

x^2+(25-3x/4)^2=25

We move all terms to the left:

x^2+(25-3x/4)^2-(25)=0

We add all the numbers together, and all the variables

x^2+(-3x/4+25)^2-25=0

We multiply all the terms by the denominator


x^2*4-3x+25)^2+(-25*4+25)^2=0

We add all the numbers together, and all the variables

x^2*4-3x+25)^2+(-75)^2=0

We add all the numbers together, and all the variables

x^2*4-3x=0

Wy multiply elements

4x^2-3x=0

a = 4; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·4·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*4}=\frac{0}{8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*4}=\frac{6}{8}
=3/4
$