((4x^2)/25)+x^2=196

Solution for ((4x^2)/25)+x^2=196 equation:

((4x^2)/25)+x^2=196

We move all terms to the left:

((4x^2)/25)+x^2-(196)=0

We get rid of parentheses

x^2+4x^2/25-196=0

We multiply all the terms by the denominator


4x^2+x^2*25-196*25=0

We add all the numbers together, and all the variables

4x^2+x^2*25-4900=0

Wy multiply elements

4x^2+25x^2-4900=0

We add all the numbers together, and all the variables

29x^2-4900=0

a = 29; b = 0; c = -4900;
Δ = b2-4ac
Δ = 02-4·29·(-4900)
Δ = 568400
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:

$\sqrt{\Delta}=\sqrt{568400}=\sqrt{19600*29}=\sqrt{19600}*\sqrt{29}=140\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-140\sqrt{29}}{2*29}=\frac{0-140\sqrt{29}}{58}
=-\frac{140\sqrt{29}}{58}
=-\frac{70\sqrt{29}}{29}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+140\sqrt{29}}{2*29}=\frac{0+140\sqrt{29}}{58}
=\frac{140\sqrt{29}}{58}
=\frac{70\sqrt{29}}{29}
$