20^2=4x^2+3x^2

Solution for 20^2=4x^2+3x^2 equation:

20^2=4x^2+3x^2

We move all terms to the left:

20^2-(4x^2+3x^2)=0

We add all the numbers together, and all the variables

-(4x^2+3x^2)+400=0

We get rid of parentheses

-4x^2-3x^2+400=0

We add all the numbers together, and all the variables

-7x^2+400=0

a = -7; b = 0; c = +400;
Δ = b2-4ac
Δ = 02-4·(-7)·400
Δ = 11200
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:

$\sqrt{\Delta}=\sqrt{11200}=\sqrt{1600*7}=\sqrt{1600}*\sqrt{7}=40\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{7}}{2*-7}=\frac{0-40\sqrt{7}}{-14}
=-\frac{40\sqrt{7}}{-14}
=-\frac{20\sqrt{7}}{-7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{7}}{2*-7}=\frac{0+40\sqrt{7}}{-14}
=\frac{40\sqrt{7}}{-14}
=\frac{20\sqrt{7}}{-7}
$