r2+3=17

Solution for r2+3=17 equation:

r2+3=17

We move all terms to the left:

r2+3-(17)=0

We add all the numbers together, and all the variables

r^2-14=0

a = 1; b = 0; c = -14;
Δ = b2-4ac
Δ = 02-4·1·(-14)
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{14}}{2*1}=\frac{0-2\sqrt{14}}{2}
=-\frac{2\sqrt{14}}{2}
=-\sqrt{14}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{14}}{2*1}=\frac{0+2\sqrt{14}}{2}
=\frac{2\sqrt{14}}{2}
=\sqrt{14}
$