4^2=1^2+x^2

Solution for 4^2=1^2+x^2 equation:

4^2=1^2+x^2

We move all terms to the left:

4^2-(1^2+x^2)=0

We add all the numbers together, and all the variables

-(1^2+x^2)+16=0

We get rid of parentheses

-x^2+16-1^2=0

We add all the numbers together, and all the variables

-1x^2+15=0

a = -1; b = 0; c = +15;
Δ = b2-4ac
Δ = 02-4·(-1)·15
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15}}{2*-1}=\frac{0-2\sqrt{15}}{-2}
=-\frac{2\sqrt{15}}{-2}
=-\frac{\sqrt{15}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15}}{2*-1}=\frac{0+2\sqrt{15}}{-2}
=\frac{2\sqrt{15}}{-2}
=\frac{\sqrt{15}}{-1}
$