r^2/3+10=19

Solution for r^2/3+10=19 equation:

r^2/3+10=19

We move all terms to the left:

r^2/3+10-(19)=0

We add all the numbers together, and all the variables

r^2/3-9=0

We multiply all the terms by the denominator


r^2-9*3=0

We add all the numbers together, and all the variables

r^2-27=0

a = 1; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·1·(-27)
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{3}}{2*1}=\frac{0-6\sqrt{3}}{2}
=-\frac{6\sqrt{3}}{2}
=-3\sqrt{3}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{3}}{2*1}=\frac{0+6\sqrt{3}}{2}
=\frac{6\sqrt{3}}{2}
=3\sqrt{3}
$