r2=(4)2+(3)2

Solution for r2=(4)2+(3)2 equation:

r2=(4)2+(3)2

We move all terms to the left:

r2-((4)2+(3)2)=0

We add all the numbers together, and all the variables

r2-74=0

We add all the numbers together, and all the variables

r^2-74=0

a = 1; b = 0; c = -74;
Δ = b2-4ac
Δ = 02-4·1·(-74)
Δ = 296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{296}=\sqrt{4*74}=\sqrt{4}*\sqrt{74}=2\sqrt{74}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{74}}{2*1}=\frac{0-2\sqrt{74}}{2}
=-\frac{2\sqrt{74}}{2}
=-\sqrt{74}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{74}}{2*1}=\frac{0+2\sqrt{74}}{2}
=\frac{2\sqrt{74}}{2}
=\sqrt{74}
$